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Mrac [35]
3 years ago
14

Work and how does it measure energy?

Physics
2 answers:
PSYCHO15rus [73]3 years ago
5 0
The answer to this question is that work is defined as the change in energy. It measures the change in energy. W = ΔE.
marshall27 [118]3 years ago
3 0
Energy and work are measured in joules, more commonly seen as a “J”
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When he drove home his car broke down it wasnt due to a problem with the wheels it was the?
Goshia [24]
It would have to be the engine
3 0
2 years ago
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An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11
Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

8 0
3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

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

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

4 0
3 years ago
The negatively charged student touches a metal tap and receives am electric shock.<br> Explain why.
svetlana [45]

Answer:

The potential difference between the student and the tap causes electrons to transfer to the tap which is earthed.

Hope this helps my sis and others

-Amelia

8 0
3 years ago
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An apple is dropped from the top of a building and hits the ground 2 seconds later. How tall is the building?
Ket [755]

Hello! :)

\large\boxed{19.6 m}

Use the kinematic equation below to solve for the height of the building:

d = \frac{1}{2} at^{2}

Where:

d = distance, or height of the building (m)

a = acceleration due to gravity (9.8 m/s²)

t = time (seconds)

Plug in the given values into the equation:

d = \frac{1}{2} (9.8)2^{2}\\\\d = \frac{1}{2} (39.2)\\\\d = 19.6 m

8 0
2 years ago
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