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Brrunno [24]
3 years ago
13

when an object moves down and does not stop which force is acting more strongly on the object, friction or gravity? explain

Physics
1 answer:
lina2011 [118]3 years ago
7 0

Gravity acts more strongly on the object.

<u>Explanation:</u>

When an object is dropped from a height, it reaches the ground despite friction acting on it because the force of gravity acting on it is stronger than the air resistance and friction. Air resistance and friction acts upward and prevents the ball from falling. However, it is negligible. The gravity acting on the object is so strong that it pulls the object towards earth with a constant acceleration called as acceleration due to gravity which has a constant value of 9.8m/s².

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5. How far can a car travel in 14 hours while going at a speed of 75 miles per hour?
Gemiola [76]

\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

Speed of the car is 75 miles per hour and it travels for 14 hours. so the distance travelled by it in the given period of time can be calculated using this formula ~

\qquad \sf  \dashrightarrow \: d = v \sdot t

\qquad \sf  \dashrightarrow \: d = 75 \times 14

\qquad \sf  \dashrightarrow \: d = 1050 \:  \: miles

so, it can travel 1050 miles in 14 hours

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2 years ago
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Dafna11 [192]
The answer is true .....
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3 years ago
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enyata [817]

Answer:

I think it would be the 3rd or 4th one.

8 0
3 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
2 years ago
three letters (JET) are placed in front of a plane mirror the image formed is in what arrangement???​
mario62 [17]

Answer:

TEJ as this is a thing you wont get

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3 years ago
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