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2 years ago

Shelby the Skater's rocket provides a forward force of

Physics

1 answer:

DochEvi [55]2 years ago

4 0

Answer:

$a=3.17\ m/s^2$

Explanation:

Given that,

Force force acting on the Skater's rocket is 220 N

Resistive force acting on it, F' = 30 N

Mass of Shelby, m = 60 kg

We need to find the acceleration of Shelby. Net force acting on Shelby is given by :

Net force = 220-30

= 190 N

The formula of net force is :

F = ma

a is acceleration of Shelby

$a=\dfrac{F}{m}\\\\a=\dfrac{190}{60}\\\\a=3.17\ m/s^2$

So, the acceleration of Shelby is $3.17\ m/s^2$

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A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

Flauer [41]

<h3>It takes 60 seconds to do the work</h3>

Solution:

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

Find the work done:

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

Find the time taken

$power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second$

Thus it takes 60 seconds to do the work

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2 years ago

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

$- mgh - F_f (s) = 0 - \frac{1}{2}mv^2$

here we know that

$F_f = \mu_k mg cos\theta$

also we know that the length of the incline is given as

$s = \frac{h}{sin\theta}$

now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

3 0

2 years ago

Which is NOT true of the Big Bang Theory?

grigory [225]

Answer:

1. it explains what's happening im the universe now

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2 years ago

An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe

nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

$f'=\frac{(v+v_o)}{(v+v_s)}f$

Where f is the actual frequency, $f'$ is the observed frequency, $v$ is the velocity of the sound waves, $v_o$ the velocity of the observer (which is negative if the observer is moving away from the source) and $v_s$ the velocity of the source (which is negative if is moving towards the observer). For this problem: