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4 years ago

A car of mass 2400kg moving at 20 m/s slams into a cement wall and comes to a halt?

Physics

1 answer:

Brut [27]4 years ago

7 0

Impulse = change in momentum

The car's momentum was (mass) x (speed)

Momentum = (2400 kg) x (20 m/s)

Momentum = 48,000 km-m/s

To completely stop the car, the impulse = -48,000 km-m/s .

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What are earths two main motions called

andre [41]

Rotation and revolution

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3 years ago

How was Bohr's atomic model different from those of previous scientists?

exis [7]

Answer:Bohr placed the electrons in distinct energy levels. Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits.

Explanation: also rutherfords was just a hypothesis while Bhor took the time to make his an experiment

4 0

3 years ago

6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw

USPshnik [31]

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$ . So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$

7 0

3 years ago

Given one mole of diamond vs one mole of graphite,

grandymaker [24]

Answer:

The pressure is $P= - 6.39*10^8Pa$

The temperature is $T =1218.63 K$

Explanation:

Generally Gibbs free energy is mathematically represented as

$G = E + PV -TS$

Where E is the enthalpy

PV is the pressure volume energy (i.e PV energy)

S is the entropy

T is the temperature

For stability to occur the Gibbs free energy must be equal to zero

Considering Diamond

So at temperature of T = 300 K

$E + PV - TS = 0$

making P the subject

$P = \frac{TS-E}{V}$

Now substituting 300 K for T , 2900 J for E ,

$3.42cm^3 = \frac{3.42}{1*10^6} = 3.42*10^{-6}m^3$ for V and $2.38 J/K$ for S

$P = \frac{(300 * 2.38)- 2900}{3.42*10^{-6}}$

$P= - 6.39*10^8Pa$

The negative sign signifies the direction of the pressure

Given that $P = 1*0^5Pa$

making T the subject

$T = \frac{PV+E}{S}$

Substituting into the equation

$T = \frac{1*10^5 * 3.42 *10^{-6}+2900}{2.38}$

$T =1218.63 K$

7 0

3 years ago

Read 2 more answers

Which statement is correct about the potential energy of a car moving down a ramp?

Paha777 [63]

Potential energy is highest when the car is released at the top of the ramp. The correct answer is option C

Potential energy is the energy possessed by a body when the body is at rest. Potential energy is at time called gravitational potential energy which as a product of mass of the body, acceleration due to gravity and the height attained by the body. That is,

P.E = mgh

When a car is moving down a ramp, the potential energy of the car can never remain the same except the car stop at a certain point.

Whenever a car is moving down a ramp, the potential energy of the car will be highest when the car is release at the top of the ramp. And lowest when the car reaches the bottom of the ramp.

The statement that is correct about the potential energy of a car moving down a ramp is:

Potential energy is highest when the car is released at the top of the ramp.

Therefore, the correct answer is option C

Learn more here: brainly.com/question/17400615

8 0

3 years ago