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egoroff_w [7]
3 years ago
13

7. Jake is bowling with a 8 kg bowling ball. He rolls the ball at 7 m/s, and it hits one stationary pin with a mass of 2 kg. The

pin goes flying forward at 12 m/s. How fast is the bowling ball now traveling?
Physics
1 answer:
Anna007 [38]3 years ago
7 0
This is a conservation of momentum problem. Initial momentum must equal final momentum. 
Momentum= mass* velecoity
I will denote Ball as b and pin as p
Initial momentum= V(initial b)* Mass(b)+ V(initial p) *mass (p)
since the pin is stationary to begin that means velocity is 0 and we can plug in for the rest
Initial momentum= 7m/s*8kg=56 mkg/s
Now for final momentum=V(final b)*mass(b)+V(final p)*mass(p) plugin
final momentum=V(final b)*8kg+12m/s*2kg
we can set final momentum= inital momentum and solve for V(final b)
56mkg/s= V(final b)*8kg+ 24mkg/s
32mkg/s=V(final b)*8kg
4m/s=V(final b)
so 4 meters/second is the answer


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Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist
andreyandreev [35.5K]

Answer:

a) It is moving at 43.15\frac{m}{s^{2}} when reaches the ground.

b) It is moving at 101.44\frac{m}{s^{2}} when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

W=K_f-K_i (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

K=\frac{mv^2}{2} (2)

with m the mass and v the velocity.

Using (2) on (1):

W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

W=Fd=wd=mgd (4)

Using (4) on (3):

mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}

v_f=43.15\frac{m}{s^{2}}

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

-mgd=-\frac{mv_i^2}{2}

Solving for initial velocity (when the boulder left the volcano):

v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}

v_i=101.44 \frac{m}{s^{2}}

3 0
2 years ago
With what maximum precision can its position be ascertained? [hint: ?p=m?v.]
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The formula for the momentum is
p = mv
As a consequence of the conservation of energy, there is also the law of conservation of momentum.
So,
Δp = Δmv = mΔv
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5 0
3 years ago
What is 3.75 x 10^-7?
pogonyaev

Answer:

Explanation:

3.75 * 10^-7

=3.75 * 1/10^7

=3.75/10000000

=3/800000000

any base which has it's power negative do it's reciprocal then the power will be positive.

8 0
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