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3 years ago

What can make an inclined plane more efficient?

Physics

1 answer:

Sophie [7]3 years ago

3 0

Answer:

Efficiency can be increase by using rollers in conjunction with the inclined plane. Wedge. The wedge is an adaptation of the inclined plane. It can be used to raise a heavy load over a short distance or to split a log

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Help, please! Thank you for your kind gesture

blagie [28]

Answer:

The last option.

Explanation:

Since you are going down, the gravitational potential energy would go down too. Thus, the gravitational potential energy decreases.

Since the gravitational potential energy is converted to kinetic energy when you move down, there is an increase in kinetic energy.

5 0

3 years ago

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Need help ASAP! please help me with the images below to score 50 points!!

ki77a [65]

It will move to the right and most likely a tiny bit down

8 0

3 years ago

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Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.8×104 m/s when at a distance

natali 33 [55]

Answer:

9.4 x 10⁴ m/s

Explanation:

In circular or elliptical orbits , the angular momentum is conserved because torque is zero.

mr₁v₁ = mr₂v₂

2.2 x 10¹¹ x 1.8 x 10⁴ = 4.2 x 10¹⁰ v₂

v₂ = 9.42 x 10⁴ m/s.

= 9.4 x 10⁴ m/s

7 0

3 years ago

A car accelerates uniformly from rest to a speed of 7.6 m/s in 4.6 s. How far does the car travel in that time?

uysha [10]

Answer:

The car will travel a distance of 17.45 meters.

Explanation:

Given:

Initial velocity $(V_i)$ = 0

Final velocity $(V_f)$ = 7.6 m/s

Time taken = 4.6 s

Acceleration = (Final velocity - Initial Velocity )/time

$a=\frac{(V_f-Vi)}{t}= \frac{7.6-0}{4.6}=1.65\ ms^-2$

We have to calculate total distance traveled by the car.

Let the distance traveled be 'd'

Equation of motion:

$d=V_i(t)+\frac{at^2}{2}$

Plugging the values.

⇒ $d=V_i(t)+\frac{at^2}{2}$

⇒ $d=0+\frac{1.65*(4.6)^2}{2}$

⇒ $d=17.45\ m$

The car will travel a distance of 17.45 meters for the above case.

4 0

3 years ago

One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R

Drupady [299]

(a) 120.8 m/s^2

The gravitational acceleration at a generic distance r from the centre of the planet is

$g=\frac{GM'}{r^2}$

where

G is the gravitational constant

M' is the mass enclosed by the spherical surface of radius r

r is the distance from the centre

For this part of the problem,

$r=R=1.17\cdot 10^6 m$

so the mass enclosed is just the mass of the core:

$M'=M=2.48\cdot 10^{24}kg$

So the gravitational acceleration is

$g=\frac{(6.67\cdot 10^{-11})(2.48\cdot 10^{24}kg)}{(1.17\cdot 10^6 m)^2}=120.8 m/s^2$

(b) 67.1 m/s^2

In this part of the problem,

$r=3R=3(1.17\cdot 10^6 m)=3.51\cdot 10^6 m$

and the mass enclosed here is the sum of the mass of the core and the mass of the shell, so

$M'=M+4M=5M=5(2.48\cdot 10^{24}kg)=1.24\cdot 10^{25}kg$

so the gravitational acceleration is

$g=\frac{(6.67\cdot 10^{-11})(1.24\cdot 10^{25}kg)}{(3.51\cdot 10^6 m)^2}=67.1 m/s^2$

8 0

4 years ago