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3 years ago

Help please the question is in the picture

Physics

1 answer:

Lemur [1.5K]3 years ago

7 0

The kinetic energy bc the energy is independent of the gravitational acceleration, and it depends only on the speed and mass. So the last answer

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Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work? Move the machine to a

Lady bird [3.3K]

Reduce the friction. Since the total energy is conserved, the only way to improve its work capacity is by reducing energy that doesnt go into work.

7 0

3 years ago

Read 2 more answers

The planet Saturn has a mass that is 95 times Earth's mass and a radius that is 9.4 times Earth's radius. What is the accelerati

ziro4ka [17]

Answer:

10.55111 m/s²

Explanation:

M = Mass of Saturn = $95\times 5.972\times 10^{24}\ kg$

r = Radius of Saturn = $9.4\times 6.371\times 10^6\ m$

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 95\times 5.972\times 10^{24}}{(9.4\times 6.371\times 10^6)^2}\\\Rightarrow g=10.55111\ m/s^2$

The acceleration due to gravity on Saturn is 10.55111 m/s²

4 0

3 years ago

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

3 years ago

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0

3 years ago

On which surface will the ball move the least distance?

Lisa [10]

Answer:

C) Frosted glass sheet

Explanation:

C) Frosted glass sheet

because it is Icy and slippery which make the ball move from its least distance

I hope you understand what it means

3 0

3 years ago

Help please the question is in the picture ​

Help please the question is in the picture