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marshall27 [118]
3 years ago
7

A bowling ball moves 18 meters every 2 seconds down the lane at a bowling alley. What is the speed of the bowling ball?

Physics
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

9m/s

Explanation:

18/2 = 9m/s

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What is a plane mirror? state the characteristics of the image formed by a plane mirror ​
Nadusha1986 [10]
A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!
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2 years ago
Why is a shadow formed
a_sh-v [17]

Answer:

Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.

Explanation:

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3 years ago
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To Mr. H's disgust, a 450-g black crow is raiding the recently-filled bird feeder. As Mr. H runs out the back door with his broo
bagirrra123 [75]

Answer:

Actually it's 2.50 m/s, sorry

Explanation:

It is solved by using momentum conservation equation

combined mass of crow and feeder = 450+670=1120 gm

let the recoil speed of feeder be v m/s

Then applying momentum conservation we get;

1120×1.5 = 670×v

v= 2.50 m/s

the speed at which the feeder initially recoils backwards = 2.50 m/s

5 0
3 years ago
If Chris throws the baseball 60 meters forward in 4.2 sec, what is the velocity of the ball? I need answer immediately
vazorg [7]

Answer:

14.3 m/s

Explanation:

velocity equation

v= d/t

v= 60/4.2

v=14.28

round it to 1 decimal place

v= 14.3m/s

5 0
2 years ago
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father
Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity \omega_o = 0

angular acceleration \alpha = 4.44 rad/s²

Using the formula:

\omega = \omega_o+ \alpha t

Making t the subject of the formula:

t= \dfrac{\omega- \omega_o}{ \alpha }

where;

\omega = 1.53 \ rad/s^2

∴

t= \dfrac{1.53-0}{4.44 }

t = 0.345 s

b)

Using the formula:

\omega ^2 = \omega _o^2 + 2 \alpha \theta

here;

\theta = angular displacement

∴

\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }

\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }

\theta =0.264 \ rad

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

x = \dfrac{0.264 \times 1}{2 \pi}

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

\tau = 270 \times 1.20 \\ \\  \tau = 324 \ Nm

However;

From the moment of inertia;

Torque( \tau) = I \alpha \\ \\  Since( I \alpha) = 324 \ Nm. \\ \\  Then; \\ \\  \alpha= \dfrac{324}{I}

given that;

I = 84.4 kg.m²

\alpha= \dfrac{324}{84.4} \\ \\  \alpha=3.84 \ rad/s^2

For re-tardation; \alpha=-3.84 \ rad/s^2

Using the equation

t= \dfrac{\omega- \omega_o}{ \alpha }

t= \dfrac{0-1.53}{ -3.84 }

t= \dfrac{1.53}{ 3.84 }

t = 0.398s

The required time it takes= 0.398s

5 0
2 years ago
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