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3 years ago

5

Cody’s car accelerates from 0m/s to 65 m/s northward in 25 seconds. What is the acceleration of the car?

1 answer:

Brut [27]3 years ago

6 0

Answer:

v=v(initial)+at

45=0+15a

a= 3m/sec²

Explanation:

edge 2020

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A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu

ira [324]

Answer:

$-20.0 kg m^2/s$

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

$I=2 kg m^2$ is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

$\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2$

7 0

3 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

3 years ago

You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle soun

mojhsa [17]

Answer:

Kindly check explanation

Explanation:

Given the following :

As train approaches ; frequency, f1 = 94Hz

As train recedes; frequency, f2 = 71Hz

Speed of sound in air ; v = 340m/s

A) speed of sound source (speed of train) = vs

From doppler effect :

As the train recedes ;

f2 = fs [v / (v + vs)] - - - - (1)

As train approaches :

f1 = fs [v / (v - vs)] ----- (2)

To find vs equate (1) and (2)

fs [v / (v - vs)] = fs [v / (v + vs)]

f1/f2 = v / (v - vs) ÷ v / (v + vs)

f1 / f2 = v / (v - vs) × (v + vs) / v

f1 / f2 = (v + vs) / (v - vs)

Let f1 / f2 = f

f = (v + vs) / (v - vs)

f (v - vs) = v + vs

fv - fvs = v + vs

fv - v = vs + fvs

v(f - 1) = vs(1 + f)

v(f - 1) / (1 + f) = vs

B)

v(f - 1) / (1 + f) = vs

f = f1 / f2 = 94/71 = 1.32 Hz

340(1.324 - 1) / (1 + 1.324) = vs

vs = 340(0.324) / 2.324

vs = 110.16 / 2.324

vs = 47.40 m/s

C.) To calculate fs, frequency of train, substitute vs into our equation.

f2 = fs [v / (v + vs)]

Following our substitikn we obtain:

fs = (2f / (f + 1))f2

D)

fs = (2f / (f + 1))f2

fs = 2(1.324) / (1.324 +1)) × 71

fs = (2.648 / 2.324) × 71

fs = 1.1394148 × 71

fs = 80.898450

fs = 80.90 Hz

3 0

3 years ago

What affects the vertical (y-component) of a projectile?

Shtirlitz [24]

A. the force of gravity

8 0

3 years ago

Read 2 more answers

A 800N student does a handstand with both hands at an angle of 15 degrees from the vertical what is the force on each hand

jeyben [28]

here we can say that net force on the student vertically upwards will be counter balance by his weight downwards

Let net force F is exerted by each hand

so here we will have

$2Fcos15 = W$

$2F cos15 = 800$

$F = \frac{800}{2 cos15}$

$F = 414.1 N$

so the force exerted by each hand will be 414.1 N

5 0

3 years ago