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3 years ago

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Determine the centripetal force on a vehicle rounding a circular curve with a radius of 80 m at a constant speed of 90 km/h if t

he vehicle's mass is 2,000 kg. (Hint: First express the velocity in meters per second)

1 answer:

quester [9]3 years ago

3 0

Mass of the vehicle = 2000 kg
Velocity of the vehicle = 90 km/hr
= 25 m/s
Radius of the curve = 80 m
Then
Centripetal force = [2000 * (25)^2]/80
= (2000 * 625)/80
= 15625 kg m/s
= 15625 Newton second
I hope that this is the answer that you were looking for and the answer has come to your desired help.

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26. A baseball traveling 38 m/s is caught by the catcher. The catcher takes 0.1 seconds to stop the ball. What is the accelerati

Furkat [3]

Answer:

The ball has an acceleration of -380 m/s², this means the ball slows down

An acceleration of -380 m/s² is the equivalent of 38.736 g's

Explanation:

Step 1: Data given

Velocity of the baseball at time t=0 = 38 m/s

At time t, the ball stops. This means v = 0

time before stops = 0.1s

Step 2: Calculate the acceleration

v= v0+at

with v= the velocity of the ball at time t = 0. v= 0

with v0 = the velocity of the ball at time t=0. v0 = 38 m/s

with a= the acceleration in m/s²

with t = time in seconds

0 = 38 + a*0.1

a = -380 m/s²

The ball has an acceleration of -380 m/s², this means the ball slows down

An acceleration of -380 m/s² is the equivalent of 38.736 g's

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What swine breed is white in color with pointed ears

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A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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3 years ago

A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

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3 years ago