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klio [65]
3 years ago
8

The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid

e it (a perfectly inelastic collision). After the impact of the bullet, the block swings up to a maximum height h. The initial speed of the bullet was 50 m/s.
Physics
1 answer:
adoni [48]3 years ago
5 0

Question

What was the initial momentum of the bullet before collision?

Answer:

10 Kg.m/s

Explanation:

Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s

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An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump
ruslelena [56]

Answer:

80%

Explanation:

Efficiency = Power output / Power input × 100 %

To calculate efficiency we need to find power output of electric pump.

We can use,

Work done = Energy change

Work done per second = Energy change per second

Work done per second = Power

Therefore, Power = Energy change per second

                              = Change in potential energy of water per second

                              =mgh / t

                              = 200× 10×6 / 10

                              = 1200 W = 1.2 kW

Now use the first equation to find efficiency,

Efficiency = \frac{1.2}{1.5} × 100%

                = 80 %

8 0
3 years ago
Which sphere of Earth is associated with plate tectonics?
Marrrta [24]

When looking at this question, we can easily start by eliminating certain answers. In the selections you've provided, you've shown atmosphere. We can easily eliminate letter A, as that makes absolutely no sense. Moving on, you also eliminate letter B, as that deals with ecosystems and whatnot. And finally, you can eliminate hydrosphere, letter C - as that's not the same. That deals with water, like oceans or rivers.

That leaves you with D) Lithosphere for your answer. The Lithosphere is the rigid part of the earth, the outermost layer, I would say. The crust / mantle. That's why it would be letter D - plate tectonics seem to have relations with the Lithosphere. The lithosphere is affected.

5 0
3 years ago
Read 2 more answers
A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe
liubo4ka [24]

Answer:

 ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

I_1=\dfrac{M+m}{2}r^2

I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2

I_1=0.75\ kg.m^2

Final mass moment of inertia

I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s

ω₂=1.20

7 0
3 years ago
Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe
deff fn [24]

Answer:

\theta_2 - \theta_1 = 156.93 degree

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

y = A sin\omega t

y = \frac{A}{5}

so now we have

\frac{A}{5} = A sin\omega t

now we have

\theta_1 = 11.53 degree

so the phase other particle in opposite direction is given as

\theta_2 = 180 - 11.53 = 168.46 degree

so we have phase difference given as

\theta_2 - \theta_1 = 168.46 - 11.53

\theta_2 - \theta_1 = 156.93 degree

7 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
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