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marishachu [46]
3 years ago
7

The car starts from a room with a constant acceleration of 5 ms-2. What path will it pass in 6 seconds and what speed will it re

ach?
Physics
1 answer:
Kobotan [32]3 years ago
4 0

Answer: The speed will be  30 m/s .

Explanation:

Given: Initial velocity of the car: u = 0 m/s

Constant Acceleration: a = 5 m/s²

Time: t= 6 seconds

To find: Final velocity(v)

Formula:  v = u+at

Substitute values in the formula, we get

v=  0+(5)(6) m/s

⇒ v= 30 m/s

i.e. Final velocity = 30 m/s

Hence, the speed will be 30 m/s .

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A 23.5 kg object travels a distance of 85 m in 30 s. Find the momentum. (Hint: you must find speed first)
VLD [36.1K]
Yup, I think you add all of them
7 0
3 years ago
Where does groundwater come from?
jek_recluse [69]
Groundwater is the water found underground in the cracks and spaces in soil, sand and rock. It is stored in and moves slowly through geologic formations of soil, sand and rocks called aquifers.


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4 0
3 years ago
Read 2 more answers
At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 m above the water. One diver runs of
Alika [10]

Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

Distance s=8.0 m

Angular speed a=1.2 rev/s

To find

Number of revolutions

Solution

From the equation of simple motion we not that

S=ut+1/2gt^{2}\\ where\\u=0\\So\\8.0m=0+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{8.0m}{0.5*9.8m/s^{2} } \\ t^{2}=1.63\\t=\sqrt{1.63} \\t=1.28s

So for the number of revolutions she makes is given as

n=a*t\\n=(1.2rev/s)(1.28s)\\n=1.532revolutions

8 0
3 years ago
For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo
Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0
3 years ago
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i
stiks02 [169]

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

<u>P.E = 0.068 J = 68 mJ</u>

4 0
3 years ago
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