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2 years ago

L

Physics

1 answer:

Sedaia [141]2 years ago

8 0

Answer:

C. 573.3 J

Explanation:

Given,

The mass of the squirrel, m = 4.5 Kg

The height of the branch where the squirrel is sitting, h = 13 m

The acceleration due to gravity, g = 9.8 m/s²

The gravitational potential energy on mass 'm' is given by the relation

P.E = mgh joules

Substituting the given values in the above equation

P.E = 4.5 Kg x 13 m x 9.8 m/s²

= 573.3 J

Therefore, the gravitational potential energy of squirrel is, P.E = 573.3 J

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A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

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2 years ago

Muscles make up ore than __ of a person body?<br>a.10<br>b.25<br>c40<br>d.75

exis [7]

40 percent of your body

8 0

3 years ago

Read 2 more answers

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

6 0

3 years ago

An object is 15 cm in front of a diverging lens with a

Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}$

$q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm$

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

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

where $h_i, h_o$ are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that $h_i >0$ , which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0

3 years ago

A difference In density between gases creates buoyancy which is the ability of an object to float. 0°C helium has a density of .

rjkz [21]

Answer:

mass = 0.18 [kg]

Explanation:

This is a classic problem where we can apply the definition of density which is equal to mass over volume.

$density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]$

mass = 0.18*1

mass = 0.18 [kg]

6 0

2 years ago