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2 years ago

5

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

ope is 9.21 × 104 N, and the mass of Earth is 5.98 × 1024 kg, what is the mass of the telescope? Round the answer to the nearest whole number

2 answers:

Luda [366]2 years ago

5 0

The answer is 11,121 kg

PtichkaEL [24]2 years ago

4 0

The gravitational attraction between the Earth and the Hubble telescope is given by:

$F=G\frac{M m}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the Hubble mass

r is the distance of the Hubble telescope from the earth's center

By rearranging the equation and substituting the numbers given by the problem, we find the mass of the telescope:

$m=\frac{F r^2}{GM}=\frac{(9.21 \cdot 10^4 N)(6.94 \cdot 10^6 m)^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(5.98 \cdot 10^{24}kg) } = 11121 kg$

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Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

Learn more about friction:

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#LearnwithBrainly

7 0

2 years ago

Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener

Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 $\frac{m^{3} }{kg}$

% Error = 20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 $\frac{m^{3} }{kg}$

% Error = - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 $\frac{J}{kg k}$

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ $\frac{V}{m}$ = $\frac{R T}{P}$

⇒ Here $\frac{V}{m}$ = Specific volume = v

⇒ v = $\frac{R T}{P}$

Put all the values in the above formula we get

⇒ v = $\frac{323}{10^{6} }$ ×81.49

⇒ v = 0.02632 $\frac{m^{3} }{kg}$

This is the specific volume by ideal gas equation.

Actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.02632 - 0.021796 = 0.004524 $\frac{m^{3} }{kg}$

% Error = $\frac{0.004524}{0.021796}$ × 100

% Error = 20.75 %

(b). From the generalized compressibility chart the value of specific volume

$\frac{V}{m}$ = v = 0.0142 $\frac{m^{3} }{kg}$

The actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.0142 - 0.021796 = $\frac{m^{3} }{kg}$

% Error = $\frac{- 0.0076}{0.021796}$ × 100

% Error = - 34.85 %

3 0

2 years ago

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900

Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

$L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}$ .

(b) Now horizontally,

$L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.$

3 0

2 years ago

The force between a 100 kg man and the Earth is 980 N. How close must two protons (1.6 X 10-19 C) be to generate the same force?

LUCKY_DIMON [66]

The distance between two protons to generate 950N of force is 0.49 X 10⁻¹⁵ m

<u>Explanation:</u>

Given:

Mass of man, m = 100 kg

Force between man and earth = 980 N

Charge of proton, q = 1.6 X 10⁻¹⁹C

Same force is generated between them

Distance between two protons, r = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2}$

where,

k is Coulomb's constant

k = 9 X 10⁹ Nm²/C²

According to the question:

$950 = k\frac{q_1q_2}{r^2}$

Solving the equation:

$950 = (9X10^9) X\frac{(1.6 X 10^-^1^9 X 1.6 X 10^-^1^9)}{r^2} \\\\r^2 = \frac{(9 X 10^9) (1.6 X 10^-^1^9 X 1.6 X 10^-^1^9)}{950} \\\\r^2 = 0.024 X 10^-^2^9\\\\r^2 = 0.24 X 10^-^3^0\\\\r = 0.49 X 10^-^1^5 m$

Therefore, the distance between two protons to generate 950N of force is 0.49 X 10⁻¹⁵ m

8 0

2 years ago

Can you check this? Samantha swam upstream for some distance in one hour. She then swam downstream the same river for the same d

ElenaW [278]

So her speed in still water will be 6mph :)

3 0

2 years ago

Read 2 more answers