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2 years ago

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A truck pulled a car of 2,350 kg a distance of 25 meters. If the car accelerates from 3 m/s to 6 m/s, whats the average force ex

erted on the car?

1 answer:

faust18 [17]2 years ago

4 0

Answer:

1,269 N

Explanation:

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A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe

galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m, y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}$

The velocity v is given by:

$\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}$

The acceleration a:

$\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}$

From the given values we get two equations:

$-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4$

We also know:

$\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}$

The magnitude of the acceleration a is:

$a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7$

The magnitude of position r is:

$r=R=68m$

Plugging in to the equation for a(t):

$\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}$

and solving for ω:

$|\omega|=0.66$

Now solve for time t:

$\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83$

Using the calculated values to compute v(t):

$\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}$

The speed of the car is:

$\sqrt{38.3^2 + (-23.5)^2} = 44.9$

The position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}$

5 0

2 years ago

Read 2 more answers

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago

Why doesn't a ball roll on forever after being kicked at a soccer game?

Naddika [18.5K]

Because of gravity and friction.

6 0

2 years ago

Read 2 more answers

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of

adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0

1 year ago

Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1

abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0

2 years ago