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lys-0071 [83]
3 years ago
13

Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

face, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood block move together as one object. How fast are they traveling?
Physics
1 answer:
lidiya [134]3 years ago
7 0

Answer:

22.22m/s

Explanation:

The momentum before a collision = momentum after collision so...

work out the momentum of the first object (the bullet)

its p = mv

0.04 kg × 300 m/s  = 0.54 kg × v

rearrange this to find v which is 0.04 x 300 = 12

so 12 = 0.54 x v

    12/0.5 = v

    v = 22.22m/s

hope this helps!

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Explanation:

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Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energ
mash [69]
The energy stored in a capacitor is given by
U= \frac{1}{2} CV^2
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.

Let's calculate the energy of the first capacitor:
U_1 =  \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J

And now the energy of the second capacitor:
U_2 =  \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J

So, the total energy stored in the two capacitors is
U=U_1 +U_2 = 1.8 \cdot 10^{-3}J
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3 years ago
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A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock
lbvjy [14]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s

Part B. Using Kinematics expression for distance, time and initial velocity

y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

8 0
3 years ago
Define habitat preservation and give at least two examples.
lisov135 [29]
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7 0
3 years ago
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are th
masha68 [24]

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

7 0
3 years ago
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