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Savatey [412]
2 years ago
14

At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear

th is 6.4x10^6 metres and the acceleration due to gravity 9.8m/s^2 on the surface of the earth.
Physics
1 answer:
Mandarinka [93]2 years ago
4 0
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
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When you're asked to calculate moments and torque, what's the main difference in calculation? Also what is a couple?
leva [86]

Answer:

a) Explanation below. b) Explanation below

Explanation:

Torque is defined as the product of a force by a radius, while momentum is defined as the product of force by a distance. Mathematically we would have

T = F * r

M = F * d

where:

T = torque = [N*m]

M = moment = [N*m]

F = force =[N]

d = distance [m]

r = radius [m]

Although they have the same units, the difference between them is the application. For the case of torque this is always applied in parts that are in rotation, such as the shafts of cars, the shafts of pumps, torque in gears and etc. While the moment can be applied to a body without the need for it to rotate.

A couple, is as its name suggests a couple of forces of equal magnitude but opposite sense and do not share a line of action. A body under the action of a couple of forces tends to rotate the body without moving it from one point to another.

7 0
3 years ago
The binding energy of a nucleus is the energy equivalent of the mass discrepancy.
Dima020 [189]

ANSWER IS NOT FALSE IT IS True

6 0
2 years ago
How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a
Luba_88 [7]

Answer:

6

Explanation:

We are given that

\theta=2.12^{\circ}

Slid width,a=0.110 mm=0.11\times 10^{-3} m

1mm=10^{-3} m

Wavelength,\lambda=582 nm=582\times 10^{-9} m

1nm=10^{-9} m

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.

asin\theta=\frac{2N+1}{2}\lambda

Using the formula

0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})

2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}

2N+1=13.98

2N=13.98-1=12.98

N=\frac{12.98}{2}\approx 6

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

5 0
3 years ago
A system of inertia 0.47 kg consists of a spring gun attached to a cart and a projectile. The system is at rest on a horizontal
Vlad [161]

Answer:

v'=0.83m/a and v=10.2m/s

Explanation:

The information that we have is:

m_1=0.050kg\\m_2=0.47kg\\\theta = 40\\h=2.2m

The maximum height of the projectile is given by the equation

h=\frac{v^2sin^2\theta}{2g}

So, rearrange for the velocity,

v=\sqrt{\frac{2gh}{sin^2\theta}}\\v=\sqrt{\frac{2*9.8*2.2}{sin^2(40)}}\\v=10.2m/s

Apply the conservation of momentum,

mvcos(\theta)=m'v'

Then rearrange the recoil speed,

v'=\frac{mvcos\theta}{m_2}\\v'=\frac{0.05*10.2*cos40}{0.47}\\v'=0.83m/s\\

4 0
3 years ago
Which of the motion variables is the same in both thex and y axis?
Softa [21]

Answer:

Acceleration (b) not sure tho

Explanation:

7 0
2 years ago
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