Mỗi giây có 2,1.1018 ion+

Explain velocity ratio. Explain and define the efficiency of a machine

Norma-Jean [14]

Velocity ratio is also defined as the ratio of a distance through which any part of a machine moves, to that which the driving part moves during the same time. An object has a mechanical advantage if it exerts a force higher than the velocity ratio.

8 0

3 years ago

One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa

Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

7 0

3 years ago

Suppose a car of mass m is moving at a constant speed v of

SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

$\arcsin(v^2/(gR))$

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

$w_r = mg\sin(\theta)$

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

$F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g$

Also, the car is making a circular motion:

$a = \frac{v^2}{R}$

Combining the equations:

$\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}$

Finally the angle is

$\arcsin(v^2/(gR))$

4 0

3 years ago

When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere

Sergeeva-Olga [200]

Lower stratosphere, this is to avoid turbulence

6 0

3 years ago

Read 2 more answers

A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

5 0

3 years ago