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Alja [10]
3 years ago
7

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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I dont know how to do question 3,4
dlinn [17]

<span>Prior to the Cambrian, organisms with hard parts (eg. shells and bones) weren't present for most the time, and it's generally such hard parts that get to fossilize. There's also been more time for Precambrian deposits to get destroyed by natural processes, </span>

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7 0
3 years ago
A 20-cm-long stick of m = 0.600 kg is lifted by a rope tied 5.0 cm from the upper end. The other end touches a smooth floor. The
Liula [17]

Answer:

https://www.chegg.com/homework-help/questions-and-answers/part-20-cm-long-stick-m-0400-kg-lifted-rope-tied-70-cm-upper-end-end-touches-smooth-floor--q81268572

3 0
2 years ago
If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object
777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0
3 years ago
Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity
Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

x_A(t) = \frac{1}{2}a t^2

where

a = 6 m/s^2 is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2

where a (acceleration) is the same as traveller A, and

u_B = 20 m/s

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get x_B=0, which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s

4 0
3 years ago
2. Jak myślisz, dlaczego w naszym słowniku bez problemu odnajdziemy następujące wyrażenia
Juliette [100K]

Answer:

..........,

Explanation:

6 0
4 years ago
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