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Sergio [31]
3 years ago
15

The speed of a moving bullet can be deter-

Physics
1 answer:
solmaris [256]3 years ago
3 0

Answer:

v=16.806\ m/s

Explanation:

<u>Rotational Motion </u>

An object is said to have rotational motion is it moves at the same distance to a fixed point called the center. The distance is called radio and the number of revolutions (or radians) it makes over time is called angular speed. Knowing the angle \theta it makes in a certain time t, the angular speed is

\displaystyle w=\frac{\theta}{t}

If we wanted to know the time it takes to describe some angle, we solve for t

\displaystyle t=\frac{\theta}{w}..........[1]

Two rotating paper disks are mounted at a distance of x= 81 cm = 0.81 m. If we knew the time the bullet take from one to the other, we could determine the speed of the bullet

\displaystyle v=\frac{x}{t}.............[2]

We can determine the time taken between the disks knowing their angular speed and the angle formed in the interval. Replacing the time from [1] into [2]

\displaystyle v=\frac{x}{\frac{\theta}{w}}

\displaystyle v=\frac{xw}{\theta}

We have the values

\theta=13.7^o=13.7/(2\pi)=2.18\ rad

w=432\ rev/min=432*2\pi /60=45.239\ rad/s

Thus

\displaystyle v=\frac{(0.81)(45.239)}{2.18}

\boxed{v=16.806\ m/s}

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Momentum = mass * speed
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7 0
3 years ago
A 0.50-kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the a
marysya [2.9K]

Complete question:

A 0.50 kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the acceleration of the frog is 4.0 meters per second^2. Calculate The magnitude of the net force exerted on the frog as it leaps.

Answer:

2.0N

Explanation:

Given that,

Mass, m of the frog = 0.5 kg

The acceleration of the frog = 4.0 m/s².

We have been asked To find,

The magnitude of the net force exerted on the frog as it leaps.

So

We calculate this using the formula below :

F = ma

When we insert the values into the formula, we have:

F = 0.5 kg × 4 m/s²

F = 2.0 N

Therefore, the magnitude of net force is 2.0 N.

5 0
4 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
A stone falls freely from rest for 8.0s what is it final velocity
NemiM [27]
Is that the full question?
7 0
3 years ago
At an amusement park, the wheelie carries passengers in a circular path of radius r = 11.2 m. If the angular speed of the wheeli
Dafna1 [17]

Answer:

(a) Tangential velocity will be 38.648 m/sec

(b) Acceleration will be 133.617m/sec^2

Explanation:

We have given radius r = 11.2 m

Angular speed \omega =0.550rev/sec=0.550\times 2\pi =3.454rad/sec

(a) We have to find the tangential velocity

We know that tangential velocity is given by  

v_t=\omega r=3.454\times 11.2=38.684m/sec

(b) We know that acceleration is given by

a=\frac{v^2}{r}=\frac{38.684^2}{11.2}=133.617m/sec^2

8 0
3 years ago
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