What type of device is a car engine?

Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a

Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

Vf = √( 2 g X)

Vf = √(2 9.8 1)

Vf = 4.43 m/s

This is the speed with which it reaches the ground

Having the final speed we can find the time

Vf = Vo + g t

t = Vf / g

t = 4.43 / 9.8

t = 0.45 s

This is the time of fall of the body to touch the ground

3 0

4 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.

klio [65]

About 40 000 km/h
Here you go:

8 0

3 years ago

Read 2 more answers

Which kind of force do you exert on an object when you pull it toward you?

Vilka [71]

It's gravity I believe? I'm not entirely sure.

7 0

4 years ago

Read 2 more answers

Add the following displacement vectors: <br> 12 m south and 15 m 55° E of N??

daser333 [38]

12m S=0m E, -12m N
15m 55d E of N = 15 sin 55, 15 cos 55 N
Sum= (15sin55)m E, (-12 + 15 cos 55)m N

4 0

3 years ago