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3 years ago

What is a sloping surface, like a ramp, that reduces the amount of force required to do work.

Physics

1 answer:

puteri [66]3 years ago

6 0

If this is about simple machines the answer is: Inclined plane

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Geosynchronous satellites orbit the Earth at a distance of 42 000 km from the Earth's center. Their angular speed at this height

frozen [14]

Answer:

Force will be 7.97 N

Explanation:

We have given that earth is at s distance of 42000 km from the earth center

So r = 42000 km = 42000000 m

Mass m = 1500 kg

Angular velocity of earth $\omega =7.29\times 10^{-5}rad/sec$

We have to find the force

We know that force is given by

$F=m\omega ^2r=1500\times (7.29\times 10^{-5})^2\times 42000000=797.1615\times 10^{-2}=7.97N$

7 0

3 years ago

An example of an object in projectile motion is

Usimov [2.4K]

Answer: a. a leaping frog

Explanation: In a projectile motion the object must have a certain path or trajectory with respect to a certain angle. For the leaping frog its velocity can be resolve into components having the final velocity at the highest point to be zero.

4 0

3 years ago

Read 2 more answers

The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing

Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

6 0

3 years ago

Can any juniors who go to Texas connections academy help me out with physics?

zimovet [89]

I'm not from that school but I can help you.

3 0

2 years ago

A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?

pychu [463]

Answer:

C. 30.6m

Explanation:

To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

$H=ut+\frac{1}{2}gt^2$

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.

Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

$H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m$

3 0

3 years ago