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MissTica
3 years ago
13

What is the weight of a 45 kg box?

Physics
2 answers:
Sladkaya [172]3 years ago
6 0

Answer:

441

Explanation:

9.8*45=441n

8_murik_8 [283]3 years ago
4 0
45 kg.Hope it helps! :)
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A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108
tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

6 0
3 years ago
Two substances, M and N, have specific heats c and 2c. if heats Q and 4Q are supɔlied to Mand N, respectively, their changes in
Ierofanga [76]

Answer:

If the mass of B is m and the temperature change is the same, the mass of B will be 2m.

Explanation:

Q = mcT

T = mc/Q

M = 4Q/2cT........... (1)

T = Q/mc

Plug this in equation 1.

M = 4Q/(2c × Q/mc)  = 4Q ÷ 2Q/m  = 4Q × m/2Q = 2m

6 0
2 years ago
In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin
Genrish500 [490]
Energy can not be created or destroyed but can change from one form to another.

example: as a roller coaster cart loses height the more speed it gains, the potential energy is transferred into kenetic energy
8 0
3 years ago
Which point or points of view could be present in a business letter?
Studentka2010 [4]

Answer:

?

Explanation:

6 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

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

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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