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netineya [11]
2 years ago
11

In which of the following examples does the object have both kinetic and potential energy? Select all that apply.

Physics
2 answers:
notsponge [240]2 years ago
6 0
I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic
Alex2 years ago
4 0
H,g,f,d, and b are the answers


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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

3 0
2 years ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and
mina [271]

Answer:

\frac{dQ}{dt}= 4312 W

Explanation:

As we know that base of the slab is given as

A = 11 \times 8

A = 88 m^2

now we know that rate of heat transfer is given as

\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)

here we know that

k = 1.4 W/m k

Also we have

x =0.20

\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)

\frac{dQ}{dt}= 4312 W

7 0
2 years ago
Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a
Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0
2 years ago
A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

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

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
2 years ago
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