What is the momentum of a 20 kg object traveling at a rate of 5 m/s

Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .

svet-max [94.6K]

The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

<h3>Speed of the satellite</h3>

v = √GM/r

where;

M is mass of Earth
G is universal gravitation constant
r is distance from center of Earth = Radius of earth + 4930 km

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

Learn more about speed here: brainly.com/question/6504879

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5 0

2 years ago

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

3 years ago

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

6 0

3 years ago

4. A metal of mass 1.55kg was heated from 300K to 320K in 6 minutes by a boiling ring of 85 W rating, calculate the specific hea

Rufina [12.5K]

Can't say anything 'bout it.....

7 0

3 years ago

1. The choice of materials for an exciting playground slide should __________.

gavmur [86]

have a large coefficient of kinetic friction between cloth and the slide material
40(0.3+0.5)=40(0.8)=32N
F-applied=Wbook/μk
When the tire is locked, the frictional force is greater but the car cannot be steered

4 0

2 years ago

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