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3 years ago

14

If a bowling ball is moving with acceleration of 15 m/s2 and has a mass of 10kg what is the force of the ball as it knocks down

pins in a bowling alley?

1 answer:

Lina20 [59]3 years ago

6 0

Answer:

150 newtons

Explanation:

F = ma

force = mass * acceleration

the mass is 10 kg, and the accelration is 15 m/s^2

10*15 = 150

the force unit that matches with meters/second and kg is newtons

F =ma

150 newtons = 10 kg * 15 m/s^2

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A boy is swinging a yo-yo with mass 0.25 kg in a circle with radius 1 m at a speed of 6 m/s. What is the tension in the string c

Paladinen [302]

Answer:

D. 9 N

Explanation:

The tension on the string is equivalent to the centripetal force.

Centripetal force is the force exerted by an object in circular motion or path towards the center of the circular path.

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3 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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3 years ago

A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?

cluponka [151]

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e $-9.8 m/s^2$ ).

The time of the ball in air is 3.2 s, so ascending time is $3.2/2=1.6 s$ .

Therefore from kinematic equation,

$v = u + gt$

Substituting the values we get,

$0= u - 9.8 (1.6)\\\\u=15.7 m/s$ , Here v = 0 at top.

Now from equation,

$h=ut+\frac{1}{2} gt^2$ , here h is the height .

So,

$h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m$ .

Thus, the ball reached at its maximum height of 12.48 m.

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