The forces on the car at the top of the loop are the normal force and the weight, they both points downward. Also the accereation at that point points downward then Newton's second law is:
[tex]-F_N-F_g=-ma[/tex]Now, since this is a circular motion the acceleration is a centripetal one and that is given as:
[tex]a=\frac{v^2}{R}[/tex]where v is the velocity and R is the radius of the circle. Plugging this in Newton's second law we have:
[tex]-F_N-F_g=-m\frac{v^2}{R}[/tex]If the car has the minimum speed to reamin in contact then it would be on the verge of losing contact , this means that Fn=0 at the toop of the loop. Then we have that the equation above:
[tex]\begin{gathered} -F_N-F_g=-m\frac{v^2}{R} \\ 0-mg=-m\frac{v^2}{R} \\ v=\sqrt[]{gR} \end{gathered}[/tex]Plugging the values of the acceleration of gravity and the radius of the loop we have:
[tex]\begin{gathered} v=\sqrt[]{(9.8)(22.7)} \\ v=14.92 \end{gathered}[/tex]Therefore the minimum velocity is 14.92 meters per second.
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.
As the object travels along the track, what is the maximum height that it reaches above point E?
Given:
Mass of object = 2 kg
At point E, the gravitational potential energy is 0.
Let's find the maximum height the object will reach above point E.
[tex]PE=mgh[/tex]Where:
m is the mass
g is acceleration due to gravity
h is the height.
The maximum height the object will reach above point E will be the height at Point C if the restriction is to the given points.
Point C is 20 m above the ground.
Therefore, the maximum height that it reaches above point E is 20 m.
ANSWER:
20 m.
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
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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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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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 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]If a 150 pound weight is on a frictionless surface, raised at an angle of 35 degrees, what is the tension in the rope that keeps it from sliding down? What is the force perpendicular to the surface?
This is the given situation.
Where m is the mass of the block and g is the acceleration due to gravity. It is given in the question, mg=150 pound=68.04 kg.
There are two components of weight. One along with x-direction and the other with negative y-direction.
x-component is
[tex]mg\sin \theta[/tex]y-component is
[tex]mg\cos \theta[/tex]Tension on the string is equal to the x-component of the weight. and the normal force,i.e. perpendicular force is equal and opposite to the y component of the weight. But tension is in opposite direction to the x-component of weight and perpendicular force is opposite to the y-component.
Therefore the tension is,
[tex]T=-mg\sin \theta=-68.04\times\sin 35^o=-39.03\text{ N}[/tex]And the normal force is,
[tex]N=mg\cos \theta=60.04\times\cos 35^o=55.74\text{ N}[/tex]Therefore the magnitude of the tension on the string is 39.03 N
And the normal force is 55.74 N
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
Obtain the potential on the x-axis at x = 0 for the following point charge distributions on the x-axis: 200 μ C at x = 20 cm, -3 00 μ C at x = 30 cm and - 400 μ C at x = 40 cm.
Potential on x axis will be 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex] in all three cases
Electric potential is defined as the amount of work needed to move a unit charge from a reference point to a specific point against the electric field.
v1 = k ( q1 / r1)
= 9 * [tex]10^{9}[/tex] * ( 200 * [tex]10^{-6}[/tex] / 20 * [tex]10^{-2}[/tex] )
= 90 * [tex]10^{5}[/tex] = 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex]
v2 = k ( q2 / r2 )
= 9 * [tex]10^{9}[/tex] * (-300 * [tex]10^{-6}[/tex] / 30 * [tex]10^{-2}[/tex] )
= 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex]
v3 = k ( q3 / r3 )
= 9 * [tex]10^{9}[/tex] * 400 * [tex]10^{-6}[/tex] / 40 * [tex]10^{-2}[/tex]
= 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex]
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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]Outline alpha, betta and gamma radiation in terms of depth of tissue penetration, ionizing effect and speed of radiation.
Penetration and Ionizing effect gamma, betta, and alpha radiation.
The speed of radiation is alpha, betta, and gamma radiation.
A gamma ray, also known as gamma radiation, is a penetrating shape of electromagnetic radiation arising from the radioactive decay of atomic nuclei. It consists of the shortest wavelength electromagnetic waves, typically shorter than the ones of X-rays.
Gamma rays have a lot of penetrating strength that several inches of dense material like lead, or maybe some feet of concrete may be required to forestall them. Gamma rays can skip absolutely through the human body; as they pass through, they are able to purpose ionizations that harm tissue.
These are a number of the most deadly radiation known. If a person happened to be close to a gamma-ray producing item, that they had been fried in an instantaneous. honestly, a gamma-ray burst should affect existence's DNA, inflicting genetic harm lengthy after the burst is over.
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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
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]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]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.
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)
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 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
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.
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.
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 spring with a spring constant of 30.0N/m Is compressed 5.00m. What is the force that the spring would apply
Force of the Spring = -(Spring Constant) x (Displacement)
so , Here we need to put the values
F = 30* 5 = 150N
Here, the Force applied is 150N. So, the correct ans is C
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
An elevator uses 100 000J of electrical energy to raise a load of 800 N through a height of 40m in a time of 20s. What fraction of the energy input was NOT transferred to the load? A) 32% B) 68% C) 80%D) 97%
W: work
F: force required to lift load; F = 800 N
d: distance lifted: d = 40 m
W = Fd = 800*40
W = 32000 J
If 100000J were exerted on a load for which only 32000J was necessary, the other 100000-32000 = 68000J was not transferred to the load.
68000/100000 = 0.68 = 68%
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:
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.
Learn more about gamma rays here:
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I need help with one of my physics question
Determine whether the statement is true or false.2 ∈ {x | x ∈ N and x is even}Is the statement true or false?❑ True❑ False
ANSWER
True
EXPLANATION
The set is the set of all the natural even numbers. The natural numbers are {1, 2, 3, 4, ...} and the natural even numbers are {2, 4, 6, 8, ...}. Therefore, it is true that 2 belongs to this set.