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julia-pushkina [17]
3 years ago
10

Convert the following : 6400km into metre

Physics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0
6400km=6400000 meters
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A car starts moving from the position of rest with uniform acceleration of 8m/s^calculate the distance travelled by it during 10
ad-work [718]

We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.

As it leaves from rest, it means that the initial speed is zero


t₁=9 min=540 s

t₂=10 min=600 s

x₁=at₁²/2=8*540²/2=4*291600=1166400 m

x₂=at₂²/2=8*600²/2=4*360000=1440000 m

Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel




8 0
3 years ago
There are eight men sitting on a couch. Three legs break and six men leave. How many legs are remaining
lutik1710 [3]

Explanation: Well, think about it if you see it on the logical end, there are 8 men that mean 16 legs. Then there are 4 legs of the couch as well that made it total of 20. If broken legs are of the men, then still the total will be 20 because the legs are still intact but 6 men left that means 12 legs are gone, if 3 legs break and 6 men leave. the number of legs remaining are 20–6*2–3=5....  so it's 5  I hope this helps! xoxo

8 0
3 years ago
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When one person was talking in a small room, the sound intensity level was 60 dB everywhere within the room. Then, there were 14
dolphi86 [110]

Answer:

E= 71dB

Explanation:

See attached file for step by step calculation

3 0
3 years ago
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
stepladder [879]

Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

5 0
4 years ago
The star Betelgeuse is 20 times more massive than the sun. What would be the likely impact on the motion of Earth if the sun wer
Nataliya [291]
<span>Orbital radius would decrease because the gravitational pull would increase.</span>
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4 years ago
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