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3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

Physics

1 answer:

Katen [24]3 years ago

8 0

Answer:

9.4 m/s

Explanation:

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy of the object.

So we can write:

$W=K_f - K_i$

where in this problem:

W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving for Kf,

$K_f = W+K_i = -36,733+66,120=29,387 J$

The kinetic energy of the car can be also written as

$K_f = \frac{1}{2}mv^2$

where:

m = 661 kg is the mass of the car

v is its final speed

Solving, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

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An unknown gas effuses 2.165 times faster than xe. what is the molar mass of the gas? show the set up and answer with unit.

lyudmila [28]

The unknown gas has a molar mass of 27.94 g/mol.

<h3>Steps</h3>

Use Graham's Law, which states that the rate of effusion of a gas is proportional to its molar mass squared.

Two different gases, (a) and (b), are compared in the following equation, which expresses this:

(R(a)/R(b)^2 = M(b)/M (a)

The effusion rate is R.

M, the molar mass

If gas (a) is the unknown gas and Xe(M = 131 g/mol) is the second gas, then the answer to your question is:

Square both sides, and you get:

(2.165)² = 131 g/mol/ M (a)

M(a)(2.165)² = 131

M(a) is equal to 131/(2.165)² = 27.94 g/mol.

Thus, the unknown gas has a molar mass of 27.94 g/mol.

<h3>How is molar mass determined?</h3>

Divide each element's atomic weight (found in the periodic table) by the quantity of that element's atoms in the compound. Add up the totals, and then follow the number with the units of grams/mole.

Learn more about Graham's law of effusion here brainly.com/question/22359712

#SPJ4

6 0

2 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.7 m and moment of inertia 890 kg⋅m

AveGali [126]

Answer:

Explanation:

Using the Condition That Initial Angular Velocity is equal to final angular velocity

total angular momentum is equal to angular momentum of the person + angular Momentum of the Platform $L_{T} = L_{plat} + L_{per}$

Note L= I ×ω

Final Angular momentum of the person is equal to the final angular momentum of the platform

Final Moment of Inertia of the person I_{per.f} = $mr^{2}$ =70×(2.7)²=510.3Kgm2

Initial Angular Momentum L_{i} = Final Angular Momentum L_{f}

I_{plat.i} ×ω_{plat.i} + I_{per.i} ×ω_{per.i} = I_{plat.f} ×ω_{plat.f} + I_{per.f} ×ω_{per.f}

890(0.85) + 0 =890(ω_{plat.f}) +510.3(ω_{plat.f})

756.5 = 1400.3 (ω_{plat.f})

(ω_{plat.f}) =0.54rad/s

7 0

3 years ago

Help pls ... will give brainlist

Ghella [55]

I, i’m pretty sure sorry if it’s wrong

6 0

2 years ago

PLEASE HELP

GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

$v^2-u^2=2ah$ , h is the maximum height and a = -g

$0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m$

Hence, it will go up to a height of 1.8 m.

4 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago