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3 years ago

Which object is usually found passing through the earths atmosphere

Physics

2 answers:

Kisachek [45]3 years ago

7 0

A. meteor which object is usually found passing through the earth's atmosphere are meteor

DanielleElmas [232]3 years ago

3 0

I believe your answer would be A. meteor

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What is Kinematics???

Komok [63]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The branch of mechanics which deals with the motion of an object without considering the cause of motion, is known as Kinematics ~

I hope it helps ~

6 0

2 years ago

Read 2 more answers

Two glass marbles (1 and 2), each supported by a nylon thread, are rubbed against a piece of silk and then are placed near a thi

iragen [17]

Answer:

d.) Marbles 1 and 2 repel each other, but no interaction occurs with marble

Explanation:

Rubbing the glass against the silk charges the glass positively.

Assuming polarization is ignored,

●Glass won't conduct electrical charge. Therefore, acting as a insulator.

●neutral objects won't distribute it's charge when affected by charge objects in proximity. Therefore, over the marble surface, the charge remain spherically symmetrical (undisturbed)

Marble 1 is positive

Marble 2 is positive

Since marble 1 and 2 are positive, repulsion will occur.

marble 3 is neutral (still assuming polarization is ignored), therefore, no interaction with marble 1 and 2

D is the answer.

4 0

3 years ago

Johnny walked to the other side of his building using the following steps: 10 meters North, 20 meters to the West, 40 meters Sou

nataly862011 [7]

<u>Answer: </u>

Distance traveled = 70 meters

Displacement = 36.06 meters

<u>Explanation: </u>

Let north be positive Y and east be positive X

10 meters north, displacement = 10 j meters

20 meters west, displacement = -20 i meters

40 meters south, displacement = -40 j meters

Total displacement = (10 j - 20 i – 40 j) meters = (- 20 i - 30 j) meters

Magnitude of displacement = $\sqrt{20^2+30^2}=\sqrt{1300}= 36.06 meters$

Distance traveled = 10+20+40 = 70 meters

6 0

3 years ago

A balloon contains 0.04m3 of air at a pressure of 1.20 x 105Pa. Calculate the pressure required to reduce its volume to 0.025m3

sdas [7]

Answer:

you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3

4 0

3 years ago

A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ( $v_{i}$ ) = 20.0 m/s

Final velocity of the car ( $v_{f}$ ) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)

Deceleration = (( $v_{f}$ ) - ( $v_{i}$ )) ÷ time (m/s²)

Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)

Deceleration = (- 20) ÷ 3 (m/s²)

Deceleration = - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

7 0

3 years ago