Explanation:
The first law of Newton's says that the velocity of an object will remain unless there is a force acting on the object, the second law of newton says that the force is proportional to the mass and the acceleration, and the third law of newton says that there is always an equal an opposite forces acting on two objects that interact.
So, the normal force can be described as a result of the third law of Newton. It means that it is equal in magnitude to the weight of the cart.
Therefore, the answer is:
C. Yes, the student is correct because the normal force is the equal and opposite reaction to the cart's weight being applied on a surface, as described in Newton's Third Law of Motion.
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.
How do I solve this problem Hint: 1. Draw the forces -rear car: weight downwards, normal force upwards, tension right, drag leftmiddle car: weight downwards, normal force upwards, tension right and left, drag leftlocomotive: weight downwards, normal force upwards, tension left, drag left, thrust/applied right2. The system is moving with a constant speed, so the forces are balanced!3. Start by just looking at the rear car. The tension to the right must balance the drag. This will also be equal to the tension on the middle car. 4. Next, just look at the middle car. The tension to the right must balance the drag and tension to the left. This will also be equal to the tension on the locomotive. 5. Now, just look at the locomotive. The thrust/applied force must balance the drag plus the tension.
The free body diagram of locomotive can be shown as,
Therefore, according to free body diagram, four types of forces are acting on locomotive.
The free body diagram of middle car can be shown as,
Therefore, there are two forces acting on the middle car.
The free body diagram of the last car is shown as,
Therefore, there are three number of forces acting on the last car.
According to free body diagram of locomotive, the tension in cable 2 is equal to net force acting on locomotive. Therefore, the tension in cable 2 is 1000 N.
According to free body diagram of middle car, the tension in cable 1 is equal to tension in cable 2 therefore, the tension in cable 1 is also 1000 N.
The net force generated by locomotive can be given as,
[tex]F=F_{drag}-F_{middle\text{ car}}-F_{\text{rear car}}[/tex]Substitute the known values,
[tex]\begin{gathered} F=1000\text{ N-200 N-200 N} \\ =600\text{ N} \end{gathered}[/tex]Thus, the net force generated by locomotive is 600 N.
There is a distance of 3.60 * 10 ^ 7 m between the Earth and a satellite. The magnitude of the gravitational force of attraction between the and the Earth is 400 N. What would be the magnitude of the gravitational force of attraction if the distance between them changed to 1.80 * 10 ^ 7 meters?
Given:
The initial distance between the earth and the satellite, R₁=3.60×10⁷ m
The initial force between the earth and the satellite, F₁=400 N
The changed distance between the earth and the satellite R₂=1.80×10⁷ m
To find:
The changed force between the earth and the satellite.
Explanation:
From Newton's gravitational law, the gravitational force between two objects is directly proportional to the product of the mass of the objects and inversely proportional to the square of the distance between them.
Let us assume that the mass of the earth is M and the mass of the satellite is m.
Thus,
[tex]\begin{gathered} F_1=\frac{GMm}{R_1^2} \\ \Rightarrow F_1R_1^2=GMm\text{ }\rightarrow\text{ \lparen i\rparen} \end{gathered}[/tex]And, the changed force is given by,
[tex]\begin{gathered} F_2=\frac{GMm}{R_2^2} \\ \Rightarrow F_2R_2^2=GMm\rightarrow\text{ \lparen ii\rparen} \end{gathered}[/tex]From equation (i) and equation (ii),
[tex]\begin{gathered} F_1R_1^2=F_2R_2^2 \\ \Rightarrow F_2=\frac{F_1R_1^2}{R_2^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} F_2=\frac{400\times\left(3.60\times10^7\right?^2}{\lparen1.80\times10^7)^2} \\ =1600\text{ N} \end{gathered}[/tex]Final answer:
The magnitude of the gravitational force of attraction between the earth and the satellite after the distance between them was changed is 1600 N.
3.00 m^3 of water is at 20.0°C.If you raise its temperature to60.0°C, by how much will itsvolume expand?WaterB = 207•10-6 0-1(Unit = m^3)
Given,
Initial volume of the water, V₁=3.00 m³
The initial temperature of the water, T₁=20.0 °C=293.15 K
The final temperature of the water, T₂=60 °C=333.15 K
From Charle's law, we have,
[tex]\frac{V_1}{T_1_{}}=\frac{V_2}{T_2}[/tex]On rearranging the above equation,
[tex]V_2=\frac{V_1T_2}{T_1}[/tex]On substituting the known values in the above equation,
[tex]V_2=\frac{3.00\times333.15}{293.15}=3.41m^3[/tex]Therefore the change in the volume is,
[tex]\Delta V=V_2-V_1[/tex]i.e.,
[tex]\Delta V=3.41-3.00=0.41m^3[/tex]Therefore, the volume of the water will expand by 0.41 m³
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]What does the lower scale read? Answer in units of N
We will have the following:
First, we are given:
*Mass of the breaker: 1.1kg
*Mass of water: 3.3 kg
*Mass of metallic alloy: 4.2kg
*Density of the alloy: 5300kg/m^3
*Density of water: 1000kg/m^3
Now, we find the volume of water displaced by the alloy:
[tex]V_{\text{w}}=4.2\operatorname{kg}\cdot\frac{m^3}{5300\operatorname{kg}}\Rightarrow V_w=\frac{21}{26500}m^3\Rightarrow V_w\approx7.92\cdot10^{-4}m^3[/tex]Then, from the reading in the hanging scale we will have the force experienced by the alloy due to the upthrust when placed in water, that is:
[tex]R=mg-\rho Vg[/tex]So:
[tex]R=(4.2\operatorname{kg})(9.8m/s^2)-(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)[/tex][tex]\Rightarrow R=33.39396226\ldots N\Rightarrow R\approx33.4N[/tex]The reading on the lower scale is due to the weight of the water in the breaker and upthrust on the scale:
[tex]R=g(m_1+m_2)+\rho Vg[/tex]Finally:
[tex]R=(9.8m/s^2)(1.1\operatorname{kg}+3.3\operatorname{kg})+(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)[/tex][tex]\Rightarrow R=50.886003774\ldots N\Rightarrow R\approx50.9N[/tex]So, the readin on the lower scale is approximately 50.9N.
A mass of 10 kg, initially at rest on a horizontal frictionless surface, is acted upon by a horizontal force of 25 N. The speed of the mass after it has moved 5.0 m is:
We will have the following:
[tex]\begin{gathered} W\ast d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \\ \\ \end{gathered}[/tex]Then, we will have that:
[tex]\begin{gathered} (25N)(5m)=\frac{1}{2}(10kg)v_f^2-\frac{1}{2}(10kg)(0m/s){}2 \\ \\ \Rightarrow125N\ast m=(5kg)v_f^2\Rightarrow v_f^2=\frac{25N\ast m}{kg} \\ \\ \Rightarrow v_f=5m/s \end{gathered}[/tex]So, the final velocity of the mass after it has moved 5 meters is 5m/s.
A 12V battery with 0.3 ohm internal resistance is connected to a 24V voltage source charging it.What is the power delivered by the voltage source?
-2880 Watts
Explanation
Step 1
Let
[tex]\begin{gathered} V=24V \\ Internal\text{ resistance of battery=}0.3\Omega \\ \end{gathered}[/tex]
Now, we need aplly the formula
[tex]I(current\text{ in circuit)=}\frac{e.m.f}{\text{total resistance}}=\frac{24\text{v}}{0.3\Omega}=80\text{ Ampers}[/tex]Step 2
now, we need to find the powe delivered, to do that, let's use this formula:
The power delivered by the voltage source is: PV = – I * V =
[tex]\begin{gathered} P=-I\cdot V \\ P=-80\text{ A}\cdot(36)V \\ P=-2880\text{W} \end{gathered}[/tex]hence the answer is -2880 Watt
I hope this helps you
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Is it possible to get More work out of a machine than you put in?
If we were able to get more work out of a machine than we put in, energy would be created in the process. According to the Law of Conservation of Energy, this is not possible.
Therefore, the answer is:
[tex]\begin{gathered} \text{ No, it is not possible to get more work} \\ \text{ out of a machine than we put in.} \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]
A hypothesis is:an educated guess about a problem or question.the unit of measurement for volume.the official result of an experiment.None of the choices are correct.
Answer:
an educated guess about a problem or question.
Explanation:
An hypothesis is usually concerned with predicting the outcome of an experiment. It is an educated or well structured guess as to what the outcome would be and it is put to test. Thus, the correct option is
an educated guess about a problem or question.
10 Isabella researched how waves travel through the ground during an earthquake.
She drew a diagram of one, called an S wave, moving through Earth's crust.
Wave motion
Based on her diagram, what kind of wave is an S wave?
A light
B sound
C longitudinal
D transverse
PLEASE REPLY ASAP
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]Your shopping cart has a mass of 63.9 kilograms. In order to accelerate the shopping cart down an aisle at 0.22 m/s^2, what force (in newtons, N) would you need to apply to the cart?Answer: ________ N
Given:
The mass of the cart is,
[tex]m=63.9\text{ kg}[/tex]The acceleration of the cart is,
[tex]a=0.22\text{ m/s}^2[/tex]To find:
The force applied to the cart
Explanation:
The force on the cart is,
[tex]\begin{gathered} F=ma \\ =63.9\times0.22 \\ =14.06\text{ N} \end{gathered}[/tex]Hence, the force is 14.06 N.
your calculation.11. What is the kinetic energy of a 2kg rabbit hopping at a speed of 1.25 m/s?
Kinetic energy (KE) = 1/2 m v^2
Where:
m= mass = 2 kg
v= velocity = 1.25 m/s
Replacing:
KE = 1/2 x 2 kg x (1.25 m/s)^2 = 1.5625 J
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
(a) How much power (in W) is dissipated in a short circuit of 225 V AC through a resistance of 0.280 Ω? W(b) What current (in A) flows? A
Let's name some variables:
V: voltage; V = 225V
R: resistance; R = 0.28 ohm
I: current
P: power
Let's use this equation first to solve for P:
P = V^2/R
Plugging in variables:
P = 225^2/0.28
P = 180803.57 W = 180.80357 kW
We can use the following to solve for I:
V = IR
225 = I*0.28
I = 803.571 A = 803571 mA
We can also use this method:
Now that we have P, we can use this to solve for I:
P = IV
Plugging in the known variables:
180803.57 = 225*I
I = 803.571 A = 803571 mA
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
the diagram shows four identical magnets that have been dipped into piles of shavings of four different metals. what can be conducted from the diagram above
We can conclude from the experiment that not all the metals can be attracted by a magnet. Since not all the metals could be picked up by the magnet
What is magnetism?A magnet is a material that is able to pick up metallic materials. This implies that the materials can be picked up by the magnets. Metals have magnetic domains in them. These domains get aligned when the magnets have been exposed to a metal.
However, the magnetic domains of all metals do not get aligned when they are exposed to a magnet. As such, not all the metals that we can see are the metals that can be attracted by a magnet.
If we look at the image, we can see that the magnet does attract the cobal and the nickel particles very well but not the aluminium and the copper shavings.
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Missing parts;
.
The diagram shows four identical magnets that have been dipped into piles of shavings of four different metals.
What conclusion can be made about a magnet’s ability to attract metals based on what is shown in this diagram?
answer choices
Magnets do not attract all metals
Magnets only attract aluminum and copper.
Magnets attract all metals.
Steel is the only metal that magnets attract.
If you pick a mass of 11.4 kg from the ground up to 0.84 m above the ground and it takes 2.3 s to do so, how much power is expended?
Power is the rate at which energy is transferred with respect to time.
When you pick a mass m from the ground up to a height h, that mass gains potential energy (U), which is given by:
[tex]U=mgh[/tex]Where g is the gravitational acceleration:
[tex]g=9.81\frac{m}{s^2}[/tex]Find the potential energy that the mass of 11.4kg gains when it is raised 0.84m:
[tex]U=(11.4\operatorname{kg})(9.81\frac{m}{s^2})(0.84m)=93.94056J[/tex]Then, divide the total potential energy by 2.3s to find the power expended in doing so:
[tex]P=\frac{U}{t}=\frac{93.94056J}{2.3s}=40.84372174\ldots W\approx41W[/tex]Therefore, the power expended to raise a mass of 11.4kg up to 0.84m above the ground in 2.3 seconds, is 41 Watts.
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]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
An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1.9 s. A passenger in the elevator is holding a
7.4 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity
is 9.8 m/s
Answer in units of N
The tension in the cord as the elevator accelerates is 76.59 N.
As the elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.9 s, if the acceleration of the elevator is a, then:
1m = 1/2 a(1.9)²
⇒ a = 2/(1.9²) = 0.55 m/s².
Mass of the bundle; m = 7.4 kg.
Acceleration due to gravity; g = 9.8 m/s².
So, the tension in the cord as the elevator accelerates is = mg + ma
= m(g+a)
= 7.4 ( 9.8 + 0.55) N.
= 76.59 N.
Hence, the tension in the cord is 76.59 N.
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A particularly scary roller coaster contains a loop-the-loop inwhich the car and rider are completely upside down. If the radiusof the loop is 22.7 m with what minimum speed must the cartraverse the loop so that the rider does not fall out while upsidedown at the top? Assume the rider is not strapped to the car.
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.
URGENT!! ILL GIVE
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.
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When you walk across a carpet during the winter you often get a shock when you touch a metal door knob. The reason is that you pick up electrons via friction between your feet and the carpet. If you pick up -97.37 microC of charge, how many electrons did you acquire ?
This is the answer tab
The total charge can be written as
[tex]Q=-97.37*10^{-6}C[/tex]And the charge of an individual electron is
[tex]q=-1.6*10^{-19}C[/tex]Thus, the number of electrons is the total charge acquired, divided by the amount of charge on each individual electron. This gives us:
[tex]n=\frac{Q}{q}=\frac{-97.37*10^{-6}}{-1.6*10^{-19}}=6.085625*10^{14}[/tex]So, we have acquired 6.085625*10^14 electrons
For the following calculation, give the answer to the correct number of significant figures.
Given,
[tex]\frac{(71.359\text{ m}-71.357\text{ m})}{(3.2\text{ s}\times3.67\text{ s})}[/tex]When calculating using the numbers with the different number of significant digits, the final answer should contain the same number of significant digits as the number with the least number of significant digits.
On simplifying the above equation,
[tex]\frac{0.002}{12}[/tex]In the above equation, the numerator (0.002) has only one significant digit.
Therefore the answer will have one significant digit
On further simplifying,
[tex]=2\times10^{-4}[/tex]Thus the correct answer is 2×10⁻⁴
How much heat is needed to increase the internal energy of a gas in a piston by 4,258 J if the gas does 801 J of work on the environment by expanding if the process
According to the First Law of Thermodynamics, the change in internal energy (ΔU) of a system is equal to the sum of the heat (Q) added to the system and the work (W) exerted over the system:
[tex]\Delta U=Q+W[/tex]The sign of the heat and the work is positive when energy is transferred to the system and negative when energy is transferred from the system to its surroundings.
In this case, the system is formed by a gas on a piston, which gains 4,258 Joules in internal energy and releases 801 Joules as work. Then:
[tex]+4258J=Q-801J[/tex]Isolate Q from the equation to find the heat needed for this process:
[tex]\Rightarrow Q=4258J+801J=5059J[/tex]Therefore, the amount of heat needed to increase the internal energy of a system by 4,258J if it releases 801 of work to the environment, is 5059J.
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)
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