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snow_tiger [21]

3 years ago

6

Your friend is bragging about his motorcycle. He claims that it can go from a stopped position to 50 miles per hour in three sec

onds. He is describing the motorcycle's
speed.

acceleration.

velocity.

direction.

2 answers:

nalin [4]3 years ago

8 0

If the motorcycle can go from 0 to 50mph in 3 seconds the person is describing the motorcycle's acceleration.

tensa zangetsu [6.8K]3 years ago

5 0

Answer:

acceleration

Explanation:

He is describing the acceleration of motorcycle.

Here the initial velocity is zero, final velocity is 50 miles / hr, time taken is 3 sec.

Convert miles per hour into metre per second.

50 miles / hr = 22.352 m/s

using first equation of motion,

v = u + a x t

$a = \left ( \frac{v - u}{t} \right )$

$a = \left ( \frac{22.352 - 0}{3} \right )$

a = 7.45 m/s^2

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A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its

bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration $\alpha =140rad/sec^2$

We know that torque $\tau =I\alpha$

And $\tau =Fr$

So $Fr=I\alpha$ , here I is moment of inertia

So $50\times 0.12=I\times 140$

$I=0.0428kgm^2$

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6 0

3 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.

zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

$heat= m s \Delta T
$

m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= $20(s_1)(70-30)=800s_1
$

Heat gained by fluid= $100(s_2)(30-20)=1000s_2
$

Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

8 0

3 years ago

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it

MrRissso [65]

Answer:

Explanation:

From the question we are told that

The moment of inertia is $I = 2.50 \ kg \cdot m^2$

The final angular speed is $w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s$

The time taken is $t = 8.0 s$

The initial angular speed is $w_i = 0\ rad/s$

Generally the average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{41.89}{8}$

=> $\alpha = 5.24 \ rad/s^2$

Generally the torque is mathematically represented as

$\tau = I * \alpha$

=> $\tau = 5.24 * 2.50$

=> $\tau = 13.09 \ N \cdot m$

5 0

3 years ago