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1 year ago

An object moving at 15 m/s slows uniformly at a rate of 2.0 m/s each second for 5.0 s. What is its final speed?

1 answer:

8 0

Hey there!

We are given ,

Acceleration, a = -2m/s^2

Initial velocity , u = 15m/s

Time , t = 5 seconds

We know that ,
V=u+at

Now , final speed ,

V = 15+(-2)(5)

V = 15-10

V = 5 m/s -> final speed

Hope this helps you dear :)
Have a good day <3

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1 year 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

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

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

the radius of earth is 6370km and of mars is 3400km. if an object weighs 200N or earth , what will be it's weight on mars. the m

4vir4ik [10]

The weight of the object on mars is about 80n

4 0

2 years ago

How to solve arctan[tan(7pi /4)] ...?

Svet_ta [14]

Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :

<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago

When a space shuttle was launched, the astronauts onboard experienced an acceleration of 32.0 m/s2 . If one of the astronauts ha

jarptica [38.1K]

Answer:

F = 1280 N

Explanation:

Given that,

Acceleration experienced by a space shuttle, a = 32 m/s²

Mass of the astronauts, m = 40 kg

We need to find the force experienced by the astronaut.

We know that the net force is equal to the product of acceleration and its mass. So,

F = ma

F = 40 kg × 32 m/s²

So,

F = 1280 N

So, 1280 N of force is experienced by the Astronaut.

8 0

2 years ago

An object moving at 15 m/s slows uniformly at a rate of 2.0 m/s each second for 5.0 s. What is its final speed?​

An object moving at 15 m/s slows uniformly at a rate of 2.0 m/s each second for 5.0 s. What is its final speed?