All categories

2 years ago

1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?

Physics

1 answer:

shepuryov [24]2 years ago

7 0

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N

You might be interested in

Why is static electricity worse in winter?

Ratling [72]

Because static electricity build up in our homes because we have the heaters on which suck moisture out of the air.

8 0

3 years ago

Which time interval has the greatest speed?

shutvik [7]

Answer:

es la 2

Explanation:

epero que te curva

4 0

3 years ago

Hooke's law describes a certain light spring of unstretched length 31.8 cm. when one end is attached to the top of a doorframe a

lubasha [3.4K]

Answer:

Explanation:

extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .

kx = mg

k is spring constant , x is extension , m is mass

k x 8.6 x 10⁻² = 7.52 x 9.8

k = 856.93 N/m

= 857 x 10⁻³ KN /m

b ) Both side is pulled by force of 188 N .

Tension in spring = 188N

kx = T

856.93 x = 188

x = .219.38 m

= 21.938 cm

= 21.9 cm .

length of spring = 31.8 + 21.9

= 53.7 cm .

6 0

2 years ago

A sheet of aluminum with a rectangular shape is attached to a vertical support by a set of hinges. Assume the sheet is uniform a

Ivenika [448]

Hey I just got home

6 0

2 years ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

2 years ago