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konstantin123 [22]
2 years ago
14

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

r when adding figures, you need to round your answer so that it reflects the figure in the problem with the least amount of significant figures to the right of the decimal if a decimal is present).
2 cm
1 cm
5.3 cm
Physics
1 answer:
Mila [183]2 years ago
5 0
Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.
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Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is
suter [353]

Answer: 0.223 m/s^{2}

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity g of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

F=G\frac{m_{E}m}{r^2}  (1)

Where:

F is the module of the force exerted between both bodies

G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}} is the gravitational constant

m_{E}=5.98(10)^{24} kg is the mass of the Earth

m are the mass of each communications satellite

r=R_{E}+h is the distance between the center of the Earth and the satellite

R_{E}=6.38(10)^{6} m is the radius of the Earth

h=3.59(10)^{7} m is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

F=mg  (2)

Combining (1) and (2):

G\frac{m_{E}m}{r^2}=mg  (3)

Isolating g:

g=\frac{G M_{E}}{r^2}  (4)

Remembering r=R_{E}+h:

g=\frac{G M_{E}}{(R_{E}+h)^2}  (5)

g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}  

Finally:

g=0.223 m/s^{2}  

5 0
3 years ago
Convert 700kg into grams<br>​
Verdich [7]

Answer:

700,000g

Explanation:

1kg = 1000 Grams (or any kg x 1000)

5 0
2 years ago
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Which scientist saw the atom as a positively charged sphere with negative particles ( electrons ) embedded within?
Zigmanuir [339]
Ernest Rutherford is the answer you are looking for my friend.
5 0
3 years ago
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A ball is thrown straight up. What will
Papessa [141]

Answer:

velocity at the top: 0 m/s

acceleration at the top: -9.8 m/s²

Explanation:

Assuming up is positive and down is negative;

The velocity of the ball at the top of its path will be 0 m/s and the acceleration will be negative.

The velocity is 0 m/s because the ball does not move at the top of its path, and it switches from a positive velocity to a negative velocity. It must go through 0 in order to go from positive to negative.

The acceleration, however, is always negative no matter where the ball is in its motion. This negative acceleration causes the ball to slow down as it reaches the top, and speed up as it reaches the bottom.

<u>Think about it:</u> If there wasn't a negative acceleration, and it was instead 0, the ball would never come back down and instead keep going in a straight line.

6 0
2 years ago
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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
3 years ago
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