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3 years ago

Please help!!! I am having a hard time and your help would really help me.

Physics

1 answer:

castortr0y [4]3 years ago

5 0

Your claim is that the water source is stopped up, the evidence is that there is supposed to be water but there is none producing into the well, there would not be water flowing into the well if it was fine, but if it is stopped up, then the water wouldn't not flow

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Which diagram models the position of a soccer ball when it has the greatest amount of gravitational potential energy?

posledela

Answer:

Explanation:

Diagram C is the correct answer, because the ball is at the point with the highest height relative to the ground, in this way all the kinetic energy has been transformed into potential energy.

We must remember that potential energy is defined as the product of mass by gravity by height

Ep = m*g*h

where:

m = mass [kg]

g = gravity acceleration [m/s²]

h = elevation [m]

So when we have a great value for h in the above equation, we will have a big value for potential energy.

8 0

3 years ago

A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.0

jolli1 [7]

Answer:

B) 0.493

Explanation:

First Crate

Formula: F=ma

120kg*1.05m/s^2= 126N

Second Crate

Formula: F=ma

3*120kg=360kg

360kg*0.71m/s^2=255.6

Find the ratio by dividing

120kg/255.6kg=0.4929

Round 0.4929=0.493

Hope this Helps!!

8 0

3 years ago

PLEASE HELP THIS IS TIMED

Harlamova29_29 [7]

Answer:

85 N backwards

Explanation:

u subtract then thats the net force and add direction

8 0

3 years ago

Read 2 more answers

Which of the following is an accurate statement?

weeeeeb [17]

An object's momentum will change if there is a non-zero net external force acting on it. This assertion is true.

<h3></h3><h3>What is momentum?</h3>

The momentum is defined as the product of mass and the velocity of the body. It is denoted by the letter P. It occurs due to the applied force. Its unit is Kg m/s².

p=mΔV

If there is a change in the velocity there must be a force acting on the object.

If an object is acted on by a non-zero net external force, its momentum will not remain constant. is an accurate statement.

Hence, option B is correct.

To learn more about the momentum refer to the link;

brainly.com/question/4956182

#SPJ1

7 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago