When analyzing the results of a scientific experiment, Kevin should base his conclusion on

The car A has a weight of 4000 lb and is traveling to the right at 3 ft/s. Meanwhile a 2000 lb car B is traveling at 6 ft/s to t

iVinArrow [24]

To solve this problem we should apply Newton's third law for which it is defined that there must be an equal reaction in the opposite direction.

From this law, if Car A generates a force on car B, that car must have opposed a force exactly the same but in the opposite direction. The car A moved to the left and generates a force of 900lb so the magnitude of the force in the car B is also 900lb but to the right (opposite direction to the first car)

The correct option is 900lb to the right.

6 0

2 years ago

A gas expands at a constant pressure of 3 atm from a volume of 0.02 cubic meters to 0.10 cubic meters. In the process it experie

Nonamiya [84]

a) $2.4\cdot 10^4 J$

For a gas transformation occuring at a constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where

p is the gas pressure

V_f is the final volume of the gas

V_i is the initial volume

For the gas in the problem,

$p=3 atm = 3\cdot 1.013\cdot 10^5 Pa = 3.039\cdot 10^5 Pa$ is the pressure

$V_i = 0.02 m^2$ is the initial volume

$V_f = 0.10 m^3$ is the final volume

Substituting,

$W=(3.039\cdot 10^5 Pa)(0.10 m^3-0.02m^2)=24312 J = 2.4\cdot 10^4 J$

b) $3.24\cdot 10^5 J$

The heat absorbed by the gas can be found by using the 1st law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed

W is the work done

Here we have

$\Delta U = 3.0\cdot 10^5 J$

$W=2.4\cdot 10^4 J$

So we can solve the equation to find Q:

$Q=\Delta U + W = 3.0\cdot 10^5 J +2.4\cdot 10^4 J = 3.24\cdot 10^5 J$

And this process is an isobaric process (=at constant pressure).

8 0

3 years ago

Newton's first law of motion is also called the law of acceleration true or false

kotykmax [81]

False. Newton first is also called the law of intertia

4 0

3 years ago

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$

$v=\sqrt{2gl}$

Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

$u=3.13\ m/s$

The initial speed of the ball $u_{1}=3.13\ m/s$

The initial speed of the block $u_{2}=0$

(a). We need to calculate the speed of the ball after collision

Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0

3 years ago

PLEASE HELP!!!!!!! MY HOMEWORK IS DUE TODAY !!!!!!!!

bogdanovich [222]

Answer:

Distance = 25000000 miles

Time = 50 hours

Explanation:

Venus is the closest planet to Earth. It is about 25 million miles away from Earth. Its precise distance depends on where both Venus and Earth are in their respective orbits

Given that

Speed V = 500000 mph

Distance d = 25 000,000 miles

Speed = distance/ time

Time = distance/speed

Time = 25000000/500000

Time = 50 hours

It will therefore take 50 hours to get to venus at that speed.

5 0

3 years ago