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inna [77]
3 years ago
5

Running with an initial velocity of 13 m/s , a horse has an average acceleration of -1.88 m/s2 . How long does it take for the h

orse to decrease its velocity to 6.3 m/s
pls :/
Physics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

3.56s

Explanation:

Ya that's the right answer.

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Where will the spacecraft be when the gravitational forces acting on it are equal?
klio [65]
It would be not be able to move yet it would be in the air

6 0
3 years ago
A car travels 200km in 3.0 hours. Determine the average velocity of the car
Masja [62]

Answer:

<h2>66.67 km/hr</h2>

Explanation:

The average velocity of the car can be found by using the formula

a =  \frac{d}{t }  \\

d is the distance

t is the time taken

From the question we have

a =  \frac{200}{3}  \\  = 66.66666...

We have the final answer as

<h3>66.67 km/hr</h3>

Hope this helps you

4 0
3 years ago
Starting at the same time, an arrow and a ball are shot horizontally with a speeds of 50 m/s and 44 m/s respectively from the to
Yuri [45]

Answer:

option D

Explanation:

given.

horizontal velocity of arrow and a ball given as  50 m/s and 44 m/s respectively from the top of a building over flat ground.

In vertical direction, they are both identical

In vertical direction the initial velocity of arrow and a ball  is 0  m/s

Their acceleration due to gravity  is same for both arrow and a ball  9.8 m/s²

they will react bottom at the same time

 time of flight is same for both

now,

In horizontal direction,

distance = speed × time

Since speed is more for arrow, it will travel more horizontal distance  at the same time.

the correct answer is option D

5 0
3 years ago
A 98 N ball is suspended from a cable so that it hangs 3.5 m above the earth. Find the mass of the ball and the
expeople1 [14]

Answer:

Yes

Explanation:

4 0
3 years ago
Observe the given figure and find the the gravitational force between m1 and m2.​
Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0
3 years ago
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