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Zolol [24]
3 years ago
8

How to find the time with only the distance or height from the ground

Physics
1 answer:
Harman [31]3 years ago
5 0
You use acceleration due to gravity
and 1/2 atsqr=d
therefore 1/2 * 9.8 * tsqr= d
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A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin
Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0
2 years ago
Which of these materials is permeable?​
kogti [31]
Correct answer is letter B. sandstone
8 0
2 years ago
A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia
Bingel [31]

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

or

v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

5 0
3 years ago
A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist
marshall27 [118]

Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= \frac{71\times10^3}{4 \times \pi\times(220)^2}

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.

3 0
2 years ago
A 4 kg bowling boll sliding to the right at 8 m/s has elastic head-on collision with another 4K bowling ball initially at rest.
MAXImum [283]
The energy would be transferred from the objects. Also do not forget, add direction to your answer.
6 0
2 years ago
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