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sladkih [1.3K]
2 years ago
14

Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

ar and apply forces to it. Use the format of "if . . . then . . . because . . .” and be sure to answer the lesson question "How can Newton's laws be experimentally verified?” specific to Newton’s second law.
Physics
2 answers:
elena55 [62]2 years ago
8 0

For e2020 this is the simple answer "If force is applied to a car, then its acceleration will change proportionally, as predicted by Newton’s second law, F = ma."

{Just did this} But I used "If the toy car has force applied to it then the force would leave no choice but to move the toy car in whichever direction it is applied because the force would move the toy car according to Newton's laws." and got it right.

Tanzania [10]2 years ago
5 0
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.

How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.

Hope this helps.
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A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together
viktelen [127]

Answer:

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v (1)

Where:

m_{A}, m_{B} - Masses of the cars, in kilograms.

v_{A}, v_{B} - Initial velocities of the cars, in meters per second.

v - Velocity of the resulting system, in meters per second.

If we know that m_{A} = 2120\,kg, v_{A} = 13.4\,\frac{m}{s }, m_{B} = 2810\,kg and v_{B} = 0\,\frac{m}{s}, then the  velocity of the resulting system:

v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}

v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}

v = 5.762\,\frac{m}{s}

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy (K), in joules, is dissipated due to work done by friction (W_{f}), in joules, that is to say:

K = W_{f} (2)

\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s

\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s (2b)

Where:

\mu - Coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s- Travelled distance, in meters.

If we know that v = 5.762\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and s = 1.97\,m, then the coefficient of friction is:

\mu = \frac{v^{2}}{2\cdot g\cdot s}

\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}

\mu = 0.859

The coefficient of friction between the cars and the road is 0.859.

6 0
2 years ago
On the moon, what would be the force of gravity acting on an object that has a mass of 7 kg?
solniwko [45]

Answer: 7 kg

Explanation:

5 0
3 years ago
How much energy must be added to a bowl of 125 popcorn kernels in order for them to reach a popping temperature of 175°C? Assume
Lisa [10]

Answer:

c) 3176 J

Explanation:

mass of 125 popcorn = 125 x .0001 = . 0125 kg

increase in temperature =  175 - 21 = 154°C

specific heat of popcorn = 1650

heat required to increase temperature

= mass of popcorn x increase in temperature x specific heat

= .0125 x 154 x 1650

=  3176.25 J

8 0
3 years ago
Calculate the missing angles in the following diagram if the index of refraction is 1.00
defon
Glad you are coming back tomorrow I am so happy to be with you guys I will
4 0
3 years ago
Water drips from the nozzle of a shower onto the floor 190 cm below. The drops fall at regular (equal) intervals of time, the fi
VARVARA [1.3K]

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

t = \sqrt{0.388} = 0.62 s

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

8 0
3 years ago
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