a)
Using conservation of energy for ball 1:
[tex]\begin{gathered} E1=E2 \\ K1+U1=K2+U2 \\ 0+m1gh=\frac{1}{2}m1v1^2 \end{gathered}[/tex]Solve for v1:
[tex]\begin{gathered} v1=\sqrt[]{2gh} \\ v1=\sqrt[]{2(9.8)(2.5)} \\ v1=7\frac{m}{s} \end{gathered}[/tex][tex]v1=v2^{\prime}-v1^{\prime}[/tex]Using conservation of momentum:
[tex]\begin{gathered} m1v1=m1(v2^{\prime}-v1^{\prime})+m2v2^{\prime} \\ v2^{\prime}=\frac{2m1v1}{(m1+m2)} \\ \end{gathered}[/tex]a)
m2 = 5 kg
[tex]v2^{\prime}=7\frac{m}{s}[/tex]so:
[tex]\begin{gathered} v1^{\prime}=7-7 \\ v1^{\prime}=0 \end{gathered}[/tex]so:
[tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}=0 \end{gathered}[/tex]If the stationary mass is 5.0 kg the height back up the ramp is 0 meters
b)
m2 = 10
[tex]v2^{\prime}=\frac{14}{3}\frac{m}{s}[/tex]so:
[tex]v1^{\prime}=-\frac{7}{3}\frac{m}{s}[/tex][tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx0.2778m \end{gathered}[/tex]If the stationary mass is 10.0 kg the height back up the ramp is 0.2778 meters
c)
m2 = 500.00kg
[tex]v2^{\prime}=\frac{14}{101}\frac{m}{s}[/tex]so:
[tex]v1^{\prime}\approx-\frac{693}{101}\frac{m}{s}[/tex][tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx2.4m \end{gathered}[/tex]If the stationary mass is 500.0 kg the height back up the ramp is 2.4 meters
Consider the direction of the net force in the problem. Which of the following is true? A) The object is pulled by a net force in 2 directions.B) You cannot determine this information from the graphs provided. C) The object is pulled by a net force in only 1 Direction.
Answer:
A) The object is pulled by a net force in 2 directions.
Explanation:
The direction of the force is given by the sign of the force. Since the force is negative in the first interval and then it is positive in the third interval, we can say that the object is pulled in two opposite directions. So, the true statement is:
A) The object is pulled by a net force in 2 directions.
A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.20 s for the ball to reach its maximum height.(a) Find the ball's initial velocity. ______m/s upward(b) Find the height it reaches. _______m
ANSWER:
(a) 31.36 m/s
(b) 50.2 m
STEP-BY-STEP EXPLANATION:
Given:
Time (t) = 3.2 s
At the maximum height, the velocity is 0
Final velocity (v) = 0 m/s
(a)
We can calculate the initial velocity as follows:
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \\ \text{ we replacing} \\ \\ -9.8=\frac{0-u}{3.2} \\ \\ u=9.8\cdot3.2 \\ \\ u=31.36\text{ m/s} \end{gathered}[/tex](b)
Now the distance is determined with the following formula:
[tex]\begin{gathered} d=ut+\frac{1}{2}at^2 \\ \\ \text{ We replacing:} \\ \\ d=(31.36)(3.2)+\frac{1}{2}(-9.8)(3.2^2) \\ \\ d=100.352-50.176 \\ \\ d=50.176\approx50.2\text{ m} \end{gathered}[/tex]Question 7
A race car traveling at 120 km/h accelerates at a rate of 0.8 m/s2 for 20 s. What is the final speed of the race car?
207.3 km/h
213.8 km/h
177.6 km/h
125.8 km/h
Answer:
177.6 km/h
Explanation:
[tex]\boxed{v=u+at}[/tex]
where:
u is initial velocity in meters per second (m/s).v is final velocity in meters per second (m/s).a is acceleration in meters per second per second (m/s²).t is time in seconds (s).Given values:
u = 120 km/ha = 0.2 m/s²t = 20 sConvert the initial velocity from kilometers per hour to meters per second by dividing the value by 3.6:
[tex]\implies u=\dfrac{120}{3.6}=\dfrac{100}{3}\; \sf m/s[/tex]
Subsist the values of u, a and t into the formula and solve for v:
[tex]\begin{aligned} \textsf{Using} \quad v&=u+at\\\\\implies v&=\dfrac{100}{3}+0.8(20)\\\\&=\dfrac{148}{3}\; \sf m/s\end{aligned}[/tex]
Convert back to km/h by multiplying the value by 3.6:
[tex]\implies v=\dfrac{148}{3} \times 3.6=177.6\; \sf km/h[/tex]
Therefore, the final speed of the race car is 177.6 km/h.
Answer:
The final speed of the race car is 177.6 km/h.
Explanation:
The final speed of the race car can be determined using the following formula:
[tex]\boxed{\rm{\:v = u + at\:}}[/tex]where:
v = final speedu = initial speed (which is 120 km/h)a = acceleration (which is 0.8 m/s^2, but needs to be converted to km/h^2)t = time (which is 20 s)To convert the acceleration from m/s² to km/h², we need to multiply it by (60 x 60) / 1000, which is equivalent to 3.6. So:
[tex]\rm{a = 0.8\: m/s^2 \times 3.6 = 2.88\: km/h^2}[/tex]Substituting the values in the formula, we get:
[tex]\rm{v = 120\: km/h + (2.88\: km/h^2 \times 20\: s)}[/tex][tex]\rm{v = 120\: km/h + 57.6\: km/h}[/tex][tex]\rm{v = 177.6\: km/h}[/tex][tex]\therefore[/tex] The final speed of the race car is 177.6 km/h.
[tex]\blue{\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]
The average speed of blood in the aorta is 0.348 m/s, and the radius of the aorta is 1.00 cm. There are about 2.00 × 109 capillaries with an average radius of 6.26 μm. What is the approximate average speed of the blood flow in the capillaries?
We are given the following information
Average speed of blood in the aorta = 0.348 m/s
Radius of the aorta = 1 cm = 0.01 m
Number of capillaries = 2.00 × 10^9
Radius of capillaries = 6.26 μm
We are asked to find the average speed of the blood flow in the capillaries.
The incoming volume flow rate of blood in the aorta must be equal to the outgoing volume flow rate in the capillaries times the number of capillaries.
[tex]Q_a=2\times10^9\cdot Q_c[/tex]The volume flow rate can be written as the product of area and speed
[tex]A_a\cdot v_a=2\times10^9\cdot A_c\cdot v_c[/tex]Recall that area is pi into the square of the radius.
[tex]\begin{gathered} \pi(0.01)^2\cdot0.348=2\times10^9\cdot\pi(6.26\times10^{-6})^2\cdot v_c \\ (0.01)^2\cdot0.348=2\times10^9\cdot(6.26\times10^{-6})^2\cdot v_c \\ v_c=\frac{(0.01)^2\cdot0.348}{2\times10^9\cdot(6.26\times10^{-6})^2} \\ v_c=0.44\times10^{-3}\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the average speed of the blood flow in the capillaries is 0.44×10^-3 m/s
A shop sign weighing 215 N hangs from the end of auniform 175-N beam as shown in (Figure 1).Find the tension in the supporting wire (at 35.0 degrees)
In order to find the tension in the wire, let's first decompose it in its vertical and horizontal components:
[tex]\begin{gathered} T_x=T\cdot\cos (35\degree) \\ T_y=T\cdot\sin (35\degree) \end{gathered}[/tex]Now, since the system is stable, the sum of vertical forces is equal to zero, so we have:
[tex]\begin{gathered} T_y-175-215=0 \\ T_y-390=0 \\ T_y=390 \\ T\cdot\sin (35\degree)=390 \\ T\cdot0.57358=390 \\ T=\frac{390}{0.57358} \\ T=679.94\text{ N} \end{gathered}[/tex]So the tension in the wire is equal to 679.94 N.
A racing car of mass 1500 kg, is accelerating at 5.0 m/s2, is experiencing a lift force of 600 N [up}, due to its streamlined shape, and grounding effects of 1000 N [down], due to air dams and spoilers. Find the driving force needed to keep the car going given that μk = 1.0.
Given data:
* The acceleration of the car is,
[tex]a=5ms^{-2}[/tex]* The mass of the car is,
[tex]m=1500\text{ kg}[/tex]* The force acting on the car in the upward direction is,
[tex]F_1=600\text{ N}[/tex]* The force acting on the car in the downward direction is,
[tex]F_2=1000\text{ N}[/tex]* The coefficient of kinetic friction is,
[tex]\mu_k=1[/tex]Solution:
The weight of the car is,
[tex]\begin{gathered} w=mg \\ w=1500\times9.8 \\ w=14700\text{ N} \end{gathered}[/tex]The normal force acting on the car is,
[tex]\begin{gathered} F_N=w+F_2-F_1 \\ F_N=14700+1000-600 \\ F_N=15100\text{ N} \end{gathered}[/tex]The frictional force acting on the car is,
[tex]\begin{gathered} F_k=\mu_kF_N \\ F_k=1\times15100 \\ F_k=15100\text{ N} \end{gathered}[/tex]According to newton's second law, the force acting on the car is,
[tex]\begin{gathered} F=ma \\ F=1500\times5.0 \\ F=7500\text{ N} \end{gathered}[/tex]The net force acting on the car in terms of the applied force and frictional force is,
[tex]\begin{gathered} F=F_a-F_k \\ F_a=F+F_k \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} F_a=7500+15100 \\ F_a=22600\text{ N} \end{gathered}[/tex]Thus, the driving force required to maintain the motion of the car is 22600 N.
Chase goes to a restaurant and the subtotal on the bill was xx dollars. A tax of 7% is applied to the bill. Chase decides to leave a tip of 15% on the entire bill (including the tax). Write an expression in terms of xx that represents the total amount that Chase paid.
From the information given, the subtotal on the bill was $x. A tax of 7% was applied on this amount. Recall, percentage is expresed in terms of 100. This means that the amount of tax charged is
7/100 * x = 0.07x
Amount of bill including tax = initial amount + tax
Amount of bill including tax = x + 0.07x = 1.07x
Chase decides to leave a tip of 15% on the entire bill (including the tax). This means that the amount paid as tip is
15/100 * 1.07x = 0.1605x
Total amount paid = amount of bill including tax + amount of tip
Total amount paid = 1.07x + 0.1605x = 1.2305x
The expression for the total amount paid in terms of x is 1.2305x
Force acting on an object or system will NOT change the momentum.
From Newton's second law, the force acting on an object is given by the rate of change of momentum.
That is,
[tex]\begin{gathered} F=\frac{dp}{dt} \\ =\frac{d(mv)}{dt} \end{gathered}[/tex]Where p is the momentum of the object, m is the mass and v is the velocity of the object.
Thus, the force acting on an object will change its momentum.
Therefore the given statement is false.
these are 5 of the hw questions that I'm struggling with, you might have to zoom in a bit.
Given data
*The given time is t = 6 seconds
*The given distance from the base of the cliif is R = 30 m
The formula for the speed is given as
[tex]\begin{gathered} R=ut \\ u=\frac{R}{t} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} u=\frac{30}{7} \\ =4.28\text{ m/s} \end{gathered}[/tex]Amplitude of damped oscillations reduces two folds during one second. Find the time of its five-fold decrease.
The time for five-fold decrease = 2.32 seconds
Explanation:The final amplitude of a damped oscillation is given as:
[tex]A=A_0e^{-kt}[/tex]The amplitude reduces two-folds during one second
That is:
t = 1 second
A = 0.5A₀
[tex]\begin{gathered} 0.5A_0=A_0e^{-kt} \\ \frac{0.5A_0}{A_0}=e^{-kt} \\ 0.5=e^{-k(1)} \\ 0.5=e^{-k} \\ \ln 0.5=-k \\ k=-\ln 0.5 \\ k=0.693 \end{gathered}[/tex]For a five-fold decrease
[tex]\begin{gathered} \frac{A_0}{5}=A_0e^{-kt} \\ 0.2A_0=A_0e^{-kt} \\ \frac{0.2A_0}{A_0}=e^{-kt} \\ 0.2=e^{-kt} \\ \ln 0.2=-kt \\ \ln 0.2=-0.693t \\ -1.609=-0.693t \\ t=\frac{-1.609}{-0.693} \\ t=2.32 \end{gathered}[/tex]The time for five-fold decrease = 2.32 seconds
How to find the compressive stress in a circular tube
The compressive stress is 177.93 MN/m²
Explanations:1 inch = 0.0254 meters
The outside diameter, D = 2.5 in
D = 2.5 x 0.0254
D = 0.0635 m
The inner diameter, d = 1.5 in
d = 1.5 x 0.0254
d = 0.0381 m
The area of circular tube is calculated as:
[tex]\begin{gathered} A\text{ = }\frac{\pi{}}{4}(D^2-d^2) \\ A\text{ = }\frac{3.142}{4}(0.0635^2-0.0381^2) \\ A\text{ = }0.002m^2 \end{gathered}[/tex]The Area of the circular tube = 0.002 m²
The compressive load = 80 kips
1 kips = 4448.22 N
The compressive load = 4448.22 x 80 N
The compressive load = 355857.6N
[tex]\begin{gathered} \text{Compressive stress = }\frac{Compressive\text{ load}}{\text{Area}} \\ \text{Compressive stress = }\frac{355857.6}{0.002} \\ \text{Compressive stress = }177928800N/m^2 \\ \text{Compressive stress = }177.93MN/m^2 \end{gathered}[/tex]Matt has a mass of 37 kg and skis down a hill with no friction or air resistance. The hill has a slop of 24°. A. What is the normal force acting on himB. What is his acceleration down hill?C. Now assume there is friction. If the coefficient of kinetic friction between him and the hill is .31, what is his acceleration down the hill?
Given:
Mass of object = 37 kg
Slope = 24 degrees
Let's solve the following questions.
A. Normal force acting on him.
To find the normal force acting on Matt, apply the formula below:
[tex]mg\cos (\theta)-N=0[/tex]Where N is the force.
m is the mass = 37 kg
g is the gravitational acceleration = 9.8 m/s^2
Θ = 24 degrees.
Thus, we have:
[tex]\begin{gathered} 37\ast9.8\cos (24)-N=0 \\ \\ 37\ast9.8(0.9135)-N=0 \\ \\ 331.25-N=0 \end{gathered}[/tex]Add N to both sides:
[tex]\begin{gathered} 331.25-N+N=0+N \\ \\ 331.25=N \\ \\ N=331.25N \end{gathered}[/tex]Therefore, the normal force acting on Matt is 331.25 N
B. Acceleration down the hill.
The acceleration down the hill will be the opposite side(side opposite the angle).
To find the acceleration down the hill, apply the formula below:
[tex]a=g\sin \theta[/tex]Thus, we have:
[tex]\begin{gathered} a=9.8\sin 24 \\ \\ a=9.8(0.4067) \\ \\ a=3.99m/s^2 \end{gathered}[/tex]Therefore, the acceleration down the hill is 3.99 m/s
C. Given:
Coefficient of friction between Matt and the hill is = 0.31
Let's find the acceleration assuming there is friction.
To find the acceleration, we have the formula:
[tex]Fg=m\ast a_g[/tex]Where:
Fg is the force due to gravity
m is the mass of Matt
Thus, we have:
[tex]Fg=37\ast3.99=147.5N[/tex]Also, let's find the force due to fricton using the formula:
[tex]F_f=uN[/tex]Where:
u is the coeficient of friction = 0.31
N is the normal force
We have:
[tex]F_f=0.31\ast147.5=45.7N[/tex]Thus, we have the formula:
[tex]F_g-F_f=m\ast a[/tex]Let's solve for a:
[tex]\begin{gathered} 147.5-45.7=37\ast a \\ \\ 101.8=37\ast a \\ \\ a=\frac{101.8}{37} \\ \\ a=2.75m/s^2 \end{gathered}[/tex]Therefore, the acceleration assuming there is friction is 2.75 m/s^2
ANSWER:
[tex]\begin{gathered} A\text{. 331.25 N} \\ \\ B.3.99m/s^2 \\ \\ \text{ C. 2.75 m/s}^2 \end{gathered}[/tex]i need physics help.The charge of an electron is - 1.6 x 10^ -19 C.Show that there are about 3 x 10^18 electrons in 5 x 10^8 nC of charge.
Given:
Charge of an electron = -1.6 x 10⁻¹⁹ C
Let's show that there are about 3 x 10¹⁸ electrons in 5 x 10⁸ nC of charge.
Using the given charge of an electron, to find the number of electrons in 5 x 10⁸ nC of charge, we have:
[tex]\frac{5*10^8*10^{-9}}{1.6*10^{-19}}=3*10^{18}[/tex]Now, let's prove this equation is true.
Simplify the left side:
[tex]\begin{gathered} \frac{5*10^8*10^{-9}}{1.6*10^{-19}} \\ \\ =\frac{5*10^{-1}}{1.6*10^{-19}} \\ \\ =3.125*10^{-18}\text{ electrons} \end{gathered}[/tex]We can see the number of electrons calculated is equivalent to the number of electrons given.
Therefore, we have shown that there are about 3 x 10¹⁸ electrons in 5 x 10⁸ nC of charge.
ANSWER:
Number of electrons = 3.125 x 10¹⁸ electrons
2 pucks on an air-hockey table. Puck A has a mass of 0.0380 kg and is moving along the x axis with a velocity of + 6.29 m/s. It takes a collision with puck B, which has a mass of 0.0760 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angle shown in the drawing, pluck A at an angle of 65 degrees to the x axis and pluc B at an angle of -37 degrees to the x axis. Find the speed of pluck A and pluck B.
Given,
The mass of puck A, m₁=0.0380 kg
The mass of puck B, m₂=0.0760 kg
The velocity of puck A before the collision, u=+6.29 m/s
The angle made by puck A with the x-axis, θ₁=65°
The angle made by puck B, θ₂=-37°
The momentum is conserved in both directions simultaneously and independently. That is, the sum x-components of the momentum before the collision and after the collision are equal. The same goes for the y-axis.
Considering the x-direction,
[tex]m_1u=m_1v_1\cos \theta_1+m_2v_2\cos \theta_2[/tex]Where v₁ is the velocity of puck A and v₂ is the velocity of puck B after the collision.
On substituting the known values,
[tex]\begin{gathered} 0.0380\times6.29=0.0380\times v_1\times\cos 65^{\circ}+0.0760\times v_2\times\cos (-37)^{\circ} \\ \Rightarrow0.24=0.016v_1+0.061v_2\text{ }\rightarrow\text{ (i)} \end{gathered}[/tex]Considering the y-direction,
[tex]0=m_1v_1\sin \theta_1+m_2v_2\sin \theta_2[/tex]On substituting the known values,
[tex]\begin{gathered} 0=0.0380\times v_1\times\sin 65^{\circ}+0.0760\times v_2\times\sin (-37)^{\circ} \\ 0=0.034v_1-0.046v_2\text{ }\rightarrow\text{ (ii)} \end{gathered}[/tex]On solving equations (i) and (ii),
[tex]\begin{gathered} v_1=3.93\text{ m/s} \\ v_2=2.90\text{ m/s} \end{gathered}[/tex]Thus the speed of pluck A is 3.93 m/s and the speed of pluck B is 2.90 m/s
Calculate the acceleration of the elevator for each 5 second interval
Given,
The weight of the student, W=500 N
Thus the mass of the student is given by,
[tex]m=\frac{W}{g}[/tex]Where g is the acceleration due to gravity,
On substituting the known values,
[tex]\begin{gathered} m=\frac{500}{9.8} \\ =51.02\text{ kg} \end{gathered}[/tex]For the first 5 seconds, the net force acting on the student is 0 N. Thus scale reads only his weight. As the net force is zero the acceleration of the elevator is also zero.
In the next 5 intervals, the net force acting on the student is F=200 N, as seen from the diagram.
Thus the acceleration of the elevator is given by the equation,
[tex]F=ma[/tex]Where a is the acceleration of the elevator.
On substituting the known values,
[tex]\begin{gathered} 200=51.02\times a \\ \Rightarrow a=\frac{200}{51.02} \\ =3.92m/s^2 \end{gathered}[/tex]Thus the acceleration in this interval is 3.92 m/s²
During the interval, 10s-15s, the net force acting on the student is zero as seen from the graph. Thus the acceleration of the elevator is also zero.
During the interval, 15 s-20 s, the net force on the student is F=-200 N as seen from the diagram.
Thus the acceleration is,
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \\ a=\frac{-200}{51.02} \\ =-3.92m/s^2 \end{gathered}[/tex]Thus the accelerating in this interval is -3.92 m/s²That is the elevator is accelerating downwards.
A5 kg box is at the top of a 2.7 m tall frictionless incline as shown in the diagram. It slides to the bottom of the incline, reaching a speed of 7.3m / s . What is the box's kinetic energy at the bottom of the incline? Whats the boxs ptential energy at the top of the incline?
ANSWER:
Kinetic energy: 133.2 J
Potential energy: 132.3 J
STEP-BY-STEP EXPLANATION:
Gven:
Mass (m) = 5 kg
Height (h) = 2.7 m
Speed (v) = 7.3 m/s
We calculate the kinetic energy and the potential energy using the respective formula in each case, as follows:
[tex]\begin{gathered} E_k=\frac{1}{2}\cdot m\cdot v^2=\frac{1}{2}(5)(7.3)^2=133.2\text{ J} \\ \\ E_p=m\cdot g\cdot h=(5)(9.8)(2.7)=132.3\text{ J} \end{gathered}[/tex]Therefore, the kinetic energy is equal to 133.2 joules and the potential energy is equal to 132.3 joules.
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BRAINLIEST!!!! AND 100 POINTS!!!!!!
Thermal energy can be converted to heat energy.
Option C is correct.
Different Energy Conversion MethodsElectrical energy is a form of chemical energy. Heat energy is a form of thermal energy. Electrical, potential, and other forms of energy can be created from mechanical energy. Heat and light can both be produced with nuclear energy.
What type of energy conversion occurs more during running?Running requires the body to transform potential energy into kinetic energy. A system's internal energy is known as its potential energy. When a system moves horizontally using kinetic energy, potential energy is consumed. Potential energy is held in the form of chemical energy within the human body.
To know more about energy conversion visit:-
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A cart on frictionless rollers approaches a smooth, curved slope h = 0.45 meters high. What minimum speed v is required for the cart to reach the top of the slope?
Given data
*The given height is h = 0.45 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
The formula for the minimum speed (v) required for the cart to reach the top of the slope is given by the conservation of energy as
[tex]\begin{gathered} U_k=U_p \\ \frac{1}{2}mv^2=mgh \\ v=\sqrt[]{2gh} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{2\times9.8\times0.45} \\ =2.96\text{ m/s} \end{gathered}[/tex]Hence, the minimum speed (v) required for the cart to reach the top of the slope is v = 2.96 m/s
The position of an object is given by the formulasx = -3t² + 2t -4 and y = -2t³ + 6t² +1A) What is the speed at t = 1sB) What is the acceleration at t = 1s
We are given that the position of an object if given by the following equations:
[tex]\begin{gathered} x=-3t^2+2t-4 \\ y=-2t^3+6t^2+1 \end{gathered}[/tex]To determine the velocity we will determine the derivative of each of the functions. For "x" we have:
[tex]x=-3t^2+2t-4[/tex]Finding the derivative with respect to time we get:
[tex]\frac{dx}{dt}=\frac{d}{dt}(-3t^2+2t-4)[/tex]Now we distribute the derivative on the left side:
[tex]\frac{dx}{dt}=\frac{d}{dt}(-3t^2)+\frac{d}{dt}(2t)-\frac{d}{dt}(4)[/tex]For the first derivative we will use the rule:
[tex]\frac{d}{dt}(at^n)=ant^{n-1}[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+\frac{d}{dt}(2t)-\frac{d}{dt}(4)[/tex]For the second derivative we use the rule:
[tex]\frac{d}{dt}(at)=a[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+2-\frac{d}{dt}(4)[/tex]For the third derivative we use the rule:
[tex]\frac{d}{dt}(a)=0[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+2[/tex]Now, since the velocity is the derivative with respect to time of the position and this is and we determine the derivative for the x-position what we have found is the velocity in the x-direction, therefore, we can write:
[tex]v_x=-6t^{}+2[/tex]Now we substitute the value of time, t = 1, we get:
[tex]\begin{gathered} v_x=-6(1)+2 \\ v_x=-6+2 \\ v_x=-4 \end{gathered}[/tex]Now we use derivate the function for "y":
[tex]\frac{dy}{dt}=\frac{d}{dt}(-2t^3+6t^2+1)[/tex]Using the same procedure as before we determine the derivative:
[tex]\frac{dy}{dt}=-6t^2+12t[/tex]This is the velocity in the y-direction:
[tex]v_y=-6t^2+12t[/tex]Now we substitute the value of t = 1:
[tex]\begin{gathered} v_y=-6(1)^2+12(1) \\ v_y=6 \end{gathered}[/tex]Now, the speed is the magnitude of the velocity, the magnitude is given by:
[tex]v=\sqrt[]{v^2_x+v^2_y}[/tex]Substituting the values we get:
[tex]v=\sqrt[]{(-4)^2+6^2}[/tex]Solving the operations:
[tex]v=7.21[/tex]Therefore, the speed is 7.21 m/s.
To determine the acceleration we will determine the derivative of the formulas for velocities:
[tex]v_x=-6t^{}+2[/tex]Now we derivate with respect to time:
[tex]\frac{dv_x}{dt}=a_x=-6[/tex]Now we use the function for the velocity in the y-direction:
[tex]\frac{dv_y}{dt}=a_y=-12t^{}+12[/tex]Now we substitute the value of t = 1:
[tex]\begin{gathered} a_y=-12(1)^{}+12 \\ a_y=0 \end{gathered}[/tex]Since the acceleration in the y-direction is zero, this means that the total acceleration is the acceleration in the x-direction, therefore, the magnitude of the acceleration is:
[tex]a=6[/tex]4. The force between two charged balls is 6.0 × 10–6 N. If the distance is doubled and the charge on one ball is doubled, what is the new force between the two charged balls? a. 3.0 × 10–6 N b. 6.0 × 10–6 N c. 3.0 × 10–3 N d. 6.0 × 10–3 N
Given:
The force between two charged balls is,
[tex]F=6.0\times10^{-6}\text{ N}[/tex]The distance between the balls is doubled and the charge on one ball is doubled.
To find:
The new force between the charged balls
Explanation:
Let, the charges are,
[tex]q_1\text{ and q}_2[/tex]The distance between the charges be d, then the force between the charged balls is,
[tex]\begin{gathered} F=k\frac{q_1q_2}{d^2}=6.0\times10^{-6}\text{ N} \\ k=Coulomb^{\prime}s\text{ constant} \end{gathered}[/tex]Now, the distance is
[tex]2d[/tex]and the first charge became,
[tex]2q_1[/tex]The new force is,
[tex]\begin{gathered} F_{new}=k\frac{2q_1q_2}{(2d)^2} \\ =k\frac{q_1q_2}{2d^2} \\ =\frac{F}{2} \\ =\frac{6.0\times10^{-6}}{2} \\ =3.0\times10^{-6}\text{ N} \end{gathered}[/tex]Hence, the new force is,
[tex]\begin{equation*} 3.0\times10^{-6}\text{ N} \end{equation*}[/tex]two vectors are defined as follows: A= (-2.2m)x and B = (1.4m)y. (a) Is the magnitude of 1.4 A greater than, less than or equal to the magnitude of 1.2B? (b) Is the x component of 1.4A greater than, less than or the same as the y component of 2.2B?
We are given the following two vectors
[tex]\begin{gathered} \vec{A}=(1.4\; m)\hat{x} \\ \vec{B}=(-2.2\; m)\hat{y} \end{gathered}[/tex](a) The magnitude of vector A (1.4 m) is less than the magnitude of vector B (2.2 m) because the absolute value of vector A (1.4 m) is less than the absolute value of vector B (-2.2 m)
[tex]\begin{gathered} |\vec{A}|=|1.4|=1.4\; m \\ |\vec{B}|=|-2.2|=2.2\; m \end{gathered}[/tex](b) The x component of vector A (1.4 m) is greater than the y component of vector B (-2.2 m) because the x component of vector A (1.4 m) is positive as compared to the y component of vector B (-2.2 m)
A speedboat increases its speed from 18.5 m/s to 30.6 m/s in a distance of 226 m. Determine the acceleration of the speedboat?
V² = U² + 2aS
30.6² = 18.5² + 2*226a
936.36 = 342.25 + 452a
936.36 - 342.25 = 452a
594.11/452
a= 1.31m/s^2
Question 4 What FITT Principle describes what kind of exercise you do? O Type O Frequency Time O Indoor
Answer:
Type
FITT is acronym that stands for Frequency, Intensity, Time, and Type.
Answer:
D) type
Explanation:
How may we need to be more intentional on viewing parenting roles differently in order to most benefit our children/students?
We can be more intentional on viewing parenting roles differently in order to most benefit our children/students through-
1. Being relational
2. Being consistent
3. Being Instructive
What is intentional Parenting?
Having a strategy and setting priorities for your time and energy is all that constitutes intentional parenting. Our daily decisions and the commitments you make are then influenced by these priorities.Being an intentional parent entails understanding that the time we spend with our children is valuable and finite, and that the choices we make about how to spend that time will have an impact for a lifetime.
Thankfully, successful parenting doesn't demand perfection in these areas of instruction and punishment, but it does call for us to be thoughtful. Therefore, we must keep the following in mind to be intentional parents:
1. Parenting with intention involves relationships.
As children grow older and their lives full with milestones and events, life becomes increasingly busy. Throughout the hectic times of school and extracurricular activities, look for methods to interact frequently. Look for chances to spend time with each of your kids alone. Find moments throughout the day to have fun, play, and converse, even if it is only for a little while.
2. Consistent (even relentless) attention is a hallmark of intentional parenting.
It calls for us to be persistent in our efforts to connect with our children, refusing to give up on potential future connections just because the current one falls short of our expectations. Parenting is not a chore for parents who do it intentionally. Every chance they have to affect their children is seen by them as a wonderful gift.
3. Parenting with intention is instructive.
Kids that grow up in intentional, relational families are frequently eager and willing to learn. Every day, search for opportunities to reinforce prior teachings or find instructive situations. Next, schedule specific, devoted times to concentrate on some larger goals.
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What is the equivalent resistance in a series circuit if there are three resistors of values 5.00, 2.00, and 3.00?10.001.030.968Ω0.1000
The answer is 10.00
in series circuit , the fo
Examine the diagram of the electromagnetic spectrum below.
Diagram of electromagnetic spectrum. Gamma waves have wavelengths of approximately 10 to the power of negative 12 meters. x-rays have wavelengths of approximately 10 to the power of 10 meters. Microwaves have wavelengths of approximately 10 to the power of negative 2 meters. Radio rays have wavelengths of approximately 10 to the power of 3 meters.
If a wave has a wavelength of 0.1 nanometer, it must be a(n)
gamma ray
microwave
radio wave
x-ray
If a wave has a wavelength of 0.1 nanometers, it must be a gamma ray.
High-frequency (or shortest wavelength) electromagnetic radiation with a large amount of energy is known as gamma rays. They can go through most materials. They can only be stopped by something solid, such a big concrete block or a lead block.
Gamma rays have frequencies above 10 Hz and wavelengths below 100 pm. They represent the most powerful type of electromagnetic radiation above 100 keV.
One of the most energetic types of light created in the universe's hotter regions are gamma rays. They are also created by radioactive material in space during supernova explosions.
Ionizing radiation, of which gamma rays are a kind, is quite harmful. Ionizing radiation is high-energy radiation that charges particles by removing electrons from their atoms.
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A student measures the voltage and current between two points in an electrical circuit. If the voltage is 110 V and the current is 0.75 A, what is the resistance, according to Ohm's law?Α. 147 ΩΒ. 109 ΩC. 0.007 ΩD. 82.50 Ω
In order to calculate the resistance, we can use the formula below (Ohm's law):
[tex]R=\frac{V}{I}[/tex]If the voltage is 110 V and the current is 0.75 ohms, the resistance will be:
[tex]\begin{gathered} R=\frac{110}{0.75}\\ \\ R=147\text{ ohms} \end{gathered}[/tex]Therefore the correct option is A.
4kg of steam is at 100°C and he is removed until there is water at 39°C how much heat is removed
1024.8 KJ Heat is removed when 4kg of steam is at 100°C and he is removed until there is water at 39°C
Mass =4 kg
ΔT=100−39=61 ∘C
Q=m×C×ΔT
C= specific heat capacity of water =4200J/(kgK)
Q=4×4200×61
=1024800 Joule.
=1024.8KJ
Heat is the amount of energy that flows from one body to another on its own as a result of their different temperatures, as opposed to internal energy, which is the sum of all the molecules' energies within an item. Although it is an energy form, heat is energy in motion. Heat is not a system's property. However, a temperature difference causes the transfer of energy as heat to take place at the molecular level.
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The freezing point of a solution with a nonvolatile solute is higher than that of the pure solvent.O TrueO False
The solute is the substance that is dissolved in the solution and it is generally smaller in quantity.
The freezing point does not depend on the kind of the solute or the size of the solute, it changes when the concentration of the solute changes.
The freezing point of the solution is lower than that of the pure solvent.
Thus, The freezing point of a solution with a nonvolatile solute is lower than that of the pure solvent.
Thus, the above statement is False.
Using the parenthesis method, convert the local value of the acceleration of gravity (g=9.798 m/s2) to the referent mph/s.
Given
Gravity = 9.798 m/s2
Procedure
Let's convert m/s2 to mph/s
1 Meters Per Second Squared (m/s2)is equal to
2,24 Miles Per Hour Per Second (mph/s)
Parethesis method:
[tex]9.798\text{ m/s2}\cdot(\frac{2.24\text{ mph/s}}{1\text{ m/s2}})=21.92\text{ mph/s}[/tex]Therefore,
9.798 Meters Per Second Squared (m/s2)is:
21,92 Miles Per Hour Per Second (mph/s)