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Alla [95]
4 years ago
7

Newton's laws of motion work well for ordinary situations on earth. However, these laws of motion do not work for all cases. Und

er what condition will Newton's second law not work?
A) For objects at extremely low speeds.
B) for objects that are extremely small.
C) For objects at extremely fast speeds.
Eliminate
D) For objects that are very far away from earth.
Physics
2 answers:
Anna11 [10]4 years ago
7 0

Answer:

C) For objects at extremely fast speeds.

Explanation:

I got it right

soldier1979 [14.2K]4 years ago
5 0
I believe the answer is C: For objects at extremely fast speeds.
Hope this helps!
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what is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?​
timurjin [86]

The initial velocity of go-kart is 2.5 m/s.

<u>Explanation:</u>

Here, the uniform acceleration of go-kart is given as 0.5 m/s². Also the time required by it to stop is also given as 5 s. As acceleration is the measure of change in velocity per unit time.

In this case, the velocity should be changed from a value to zero to come to rest. So the initial velocity will be positive value and final velocity is zero.

As we know the values of acceleration, final velocity and time, the initial velocity can be easily determined as follows.

Acceleration = \frac{Final velocity -Initial velocity}{Time}

Since, final velocity is zero, acceleration is 0.5 m/s² and time is 5 s, then,

        -0.5=\frac{-\text {Initial velocity}}{5}

        Initial velocity = 0.5 × 5 = 2.5 m/s.

So the initial velocity of go-kart is 2.5 m/s.

8 0
3 years ago
Which of the following planets helped astronomers locate another planet?
worty [1.4K]

Answer: Pluto,Mercury

Explanation:

6 0
3 years ago
In order to determine the velocity, you must know
Naddik [55]
You need to know the speed and direction of object
5 0
3 years ago
Read 2 more answers
Please help me and answer these questions.
jeka57 [31]

Answer:

Answer 2 is iron.

Answer 3 is magnetic induction.

Explanation:

I hope it's helpful!

3 0
3 years ago
A rock thrown vertically upward from the surface of the moon at a velocity of 32​m/sec reaches a height of sequals32 t minus 0.8
a_sh-v [17]

Answer:

A) v(t) = (32 - 1.6t) m/s

a(t) = -1.6 m/s²

B) t = 20 seconds

C) 320 m

D) t = 5.85 seconds going up and 34.14 seconds going down

E) 40 seconds

Explanation:

Height is given by the equation;

S(t) = 32t - 0.8t²

A) Velocity after time t is gotten from first derivative of the distance.

Thus;

v(t) = dS/dt = 32 - 1.6t

Acceleration at time t is gotten from derivative of the velocity.

Thus;

a(t) = d²S/dt² = -1.6 m/s²

B) At highest point, velocity is zero.

Thus;

32 - 1.6t = 0

1.6t = 32

t = 32/1.6

t = 20 seconds

C) To find how high the rock goes, it means we are looking for maximum height.

This will be at t = 20 seconds.

Thus;

S(20) = 32(20) - 0.8(20)²

S(20) = 640 - 320

S(20) = 320 m

D) we want to find the time it will take to reach half its maximum height.

since maximum height is 320 m, then half the maximum height is; S_½ = 320/2 = 160

Thus;

160 = 32t - 0.8t²

0.8t² - 32t + 160 = 0

Using quadratic formula, we will get;

t = 5.85 seconds going up and 34.14 seconds going down

E) time the rock is aloft = twice the time it took to reach maximum height.

Thus; t_aloft = 2 × 20 = 40 seconds

8 0
3 years ago
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