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Roman55 [17]
3 years ago
12

a 2500 kg car going 35 m/a hits a 4000kg truck that is sitting still. What is the velocity of the truck if the car transfers all

of its momentum to the truck?
Physics
1 answer:
lianna [129]3 years ago
7 0

Answer:

Velocity of truck is 21.875 m/s.

Explanation:

Given:

Mass of the car (m) = 2500 kg

Initial speed of the car (u) = 35 m/s

Mass of the truck (M) = 4000 kg

Initial speed of the truck (U) = 0 m/s (Rest)

As per question, the total momentum of the car gets transferred to the truck.

Therefore, the final momentum of the car is 0.

Momentum is given as the product of mass and velocity.

So, if momentum becomes zero means the velocity becomes 0.

Therefore, final velocity of the car (v) = 0 m/s

Let the final velocity of truck be 'V' m/s.

Now, during collision, the total momentum remains conserved.

Initial momentum  = Final momentum.

Initial momentum of car + Initial momentum of truck = Final momentum of car + Final momentum of truck.

⇒ mu+MU=mv+MV\\\\2500\times 35+4000\times 0=2500\times 0+4000\times V\\\\87500+0=0+4000V\\\\4000V=87500\\\\V=\frac{87500}{4000}\\\\V=21.875\ m/s

Therefore, the velocity of the truck is 21.875 m/s.

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What happens when a star exhausts its core hydrogen supply?
Allushta [10]

Answer:

It becomes a giant or supergiant.

Explanation:

Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.

6 0
4 years ago
Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,
Leviafan [203]

Answer:

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

Explanation:

As we know that the charge per unit length of the long cylinder is given as

\lambda = \frac{q}{L}

here we know that the electric field between two cylinders is given by

E = \frac{2k\lambda}{r}

now we know that electric potential and electric field is related to each other as

\Delta V = - \int E.dr

\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr

\Delta V = -2k \lambda ln(\frac{b}{a})

\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

7 0
3 years ago
A roller coaster travels down a 120 m track in 12.5 seconds how fast does the roller coaster go
KIM [24]

Answer:

9.6m/s

Explanation:

Using the equation S=d/t where s=speed, d=distance, and t=time

plug in the known variables

S=120m/12.5s

S=9.6m/s

4 0
3 years ago
Which of the following is most likely to be an observation made by a physiologist
Naya [18.7K]
Do you have the answer choices ?
3 0
4 years ago
3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
3 years ago
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