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3 years ago

7

A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 190m. Find the the acceleration of t

he car.

1 answer:

Mama L [17]3 years ago

4 0

Answer:

a = 7.75 [m/²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$x=x_{0} +v_{o} *t + (\frac{1}{2})*a*t^{2}$

where:

x = final distance = 190 [m]

Xo = initial distance = 0

Vo = initial velocity = 0 (car starts from the rest)

a = acceleration [m/s²]

t = time = 7 [s]

190 = 0 + (0*7) + 0.5*a*(7²)

190 = 0.5*49*a

a = 7.75 [m/²]

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vova2212 [387]

<span>How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?
</span>Answer:

85 Joule (approx)

Explanation:

Potential energy at highest point

<span><span>P.E = mgh = 7.2 kg × 9.8 m/s2</span>×1.2 m≈85 Joule</span>

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4 years ago

A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the proce

Grace [21]

The angular acceleration of the wheel approximately <u>-0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s</u>.
It need approximately <u>14.474 s</u> to come to rest.

<h2>Introduction</h2>

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

1 rotation = 2π rad
1 rps = 2π rad/s
1 rpm = $\sf{\frac{1}{60} \: rps}$ = $\sf{\frac{1}{30}\pi \: rad/s}$

<h2>Formula Used</h2>

The following equations apply to proportionally changes circular motion:

<h3>Relationship between Angular Acceleration and Change of Angular Velocity </h3>

$\boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}$

With the following conditions:

$\sf{\omega_t}$ = final angular velocity (rad/s)
$\sf{\omega_0}$ = initial angular velocity (rad/s)
$\sf{\alpha}$ = angular acceleration (rad/s²)
t = interval of the time (s)

<h3>Relationship between Angular Acceleration and Change of $\sf{\theta}$ (Angle of Rotation) </h3>

$\boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}$

Or

$\boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}$

With the following condition :

$\sf{\theta}$ = change of the sudut (rad)
$\sf{\alpha}$ = angular acceleration (rad/s²)
t = interval of the time (s)
$\sf{\omega_t}$ = final angular velocity (rad/s)
$\sf{\omega_0}$ = initial angular velocity (rad/s)

<h2>Problem Solving</h2>

We know that :

$\sf{\omega_t}$ = final angular velocity = 0 rad/s >> see in the sentence "in the process of coming to rest."
$\sf{\omega_0}$ = initial angular velocity = 11 rad/s
$\sf{\theta}$ = change of the sudut = 80.0 rad

What was asked :

$\sf{\alpha}$ = angular acceleration = ... rad/s²
t = interval of the time = ... s

Step by step :

$\sf{\alpha}$ = ... rad/s²

$\sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}$

$\sf{0^2= (11)^2 + 2 \times \alpha \times 80}$

$\sf{0 = 121 + 160 \alpha}$

$\sf{-160 \alpha = 121}$

$\sf{\alpha = \frac{121}{-160}}$

$\sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}$

t = ... s

$\sf{\alpha = \frac{\omega_0 - \omega_t}{t}}$

$\sf{-0.76 = \frac{0 - 11}{t}}$

$\sf{-0.76t = -11}$

$\sf{t = \frac{- 11}{-0.76}}$

$\boxed{\sf{t \approx 14.474 \: s}}$

<h3>Conclusion</h3>

So :

The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
It need approximately 14.474 s to come to rest.

5 0

2 years ago

1. A buoyant force of 790 N lifts a 214 kg sinking boat.What is the boat's<br> net acceleration?

enot [183]

Answer:

The net acceleration of the boat is approximately 6.12 m/s² downwards

Explanation:

The buoyant or lifting force applied to the boat = 790 N

The mass of the boat lifted by the buoyant force = 214 kg

The force applied to a body is defined as the product of the mass and the acceleration of the body. Therefore, the buoyant force, F, acting on the boat can be presented as follows;

Fₐ = F - W

The weight of the boat = 214 × 9.81 = 2099.34 N

Therefore;

Fₐ = 790 - 2099.34 = -1309.34 N

Fₐ = Mass of the boat × The acceleration of the boat

Given that the buoyant force, Fₐ, is the net force acting on the boat, we have;

F = Mass of the boat × The net acceleration of the boat

F = -1309.34 N = 214 kg × The net acceleration of the boat

∴ The net acceleration of the boat = -1309.34 N/(214 kg) ≈ -6.12 m/s²

The net acceleration of the boat ≈ 6.12 m/s² downwards

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3 years ago

A phase change is when a substance changes from one state of mind to nother because of the adding or removal of thermal energy

MrRissso [65]

Answer:

true it all changes

Explanation:

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3 years ago

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JulsSmile [24]

I’ll start you off

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3 years ago