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mr_godi [17]
3 years ago
15

A block with a mass of 1kg moving at a velocity of 3m/s collides and sticks to a block of mass of 4kg initially at rest. What is

their velocity after the collision?
Physics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:using Newton third law

Let initial velocity of block be u1=3m/s

Mass of moving block m1 =1kg

Final velocity of block =V

Mass of stationary block m2= 4kg

Since they stick together, their final velocity will be the same.

m1u1 + m2u2=(m1+m2)v

(1*3)+(0*4)=(1+4)v

3=5v

Divide both sides by 5

V=0.6

Final velocity is 0.6m/s

Explanation:

You might be interested in
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation
RSB [31]

Answer:

b)  the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

4 0
3 years ago
Two spherical objects have masses 447 kg and 285 kg. Their centers are separated by a distance of 27 m. Find the gravitational a
jonny [76]

Answer:

1.2 * 10' -8N (Check attachment

Explanation:

Check attachment

5 0
3 years ago
When sitting in the tree the cat has a total of 1375J in its Gravitational potential store. What is the maximum amount of energy
Murrr4er [49]

Answer:

1375J

Explanation:

The gravitational potential/potential energy of the at the top of the tree which is the energy by virtue of its position.

P.E = mgh

mass = m

Acceleration due to gravity = g

height = h

At the top of the tree, the value of h (height) is high resulting in the gravitational potential. When the cat lands on the ground, the value of h is zero, the the gravitational potential would be zero and all the potential energy have been converted to other forms of energy.

Therefore, the total gravitational potential store is equal to the maximum amount of energy that can be transferred which is equal to 1375J.

4 0
3 years ago
Neutrons, protons, electrons are known as subatomic particles<br> explain why that is?
Alekssandra [29.7K]
<span>Particles that are smaller than the atom are called subatomic particles. The three main subatomic particles that form an atom are protons, neutrons, and electrons. The center of the atom is called the nucleus.</span>
5 0
3 years ago
Read 2 more answers
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