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Nookie1986 [14]
3 years ago
11

Which describes how fast-moving particles assist in transferring thermal energy to a cup after hot tea is poured into it?

Physics
2 answers:
enyata [817]3 years ago
5 0

Answer:

They collide with slow moving particles (B)

Explanation:

alexira [117]3 years ago
4 0

Answer:

They collide with slow moving particles (B)

Explanation:

Thermal energy is transferred through particles. Particles cannot be replaced, so both A and D are out. In order for thermal energy to be transferred, you must have 2 different substances with different temperatures, so C is out. That leaves you with A. I also took the test on Edge.

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The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
2 years ago
When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.
SpyIntel [72]

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by \tau =F\times d=45.5\times 0.0245=1.11475N-m

So torque on the rocket will be 1.11475 N -m

3 0
4 years ago
A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin
Strike441 [17]

Answer:

This question can be answered by using conversation of energy.

K_1 + U_1 = K_2 + U_2

\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2

\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s

Explanation:

Note that we take K_2 = 0 because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

7 0
4 years ago
Help !!
Vaselesa [24]
The answer is ; 6cm

Hope this helps!
Please give Brainliest!

This is because of the diagram below:

6 0
3 years ago
A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th
Hitman42 [59]

Answer:

Explanation:

Given

inclination \theta =30^{\circ}

initial speed u=20\ m/s

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

t_1=\frac{2u\sin \theta }{g}

t_1=\frac{2\times 20\times \sin 30}{10}

t_1=2\ s

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

h=u_yt+\frac{1}{2}a_yt^2

where, h=height

u_y=vertical velocity

a_y=vertical acceleration

t_0=time

45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2

t_0^2=\frac{70}{9.8}

t_0=2.64\ s

thus total time time required is t=t_0+t_1=2.64+2=4.64\ s

vertical velocity just before hitting

v_y=\sqrt{u_y^2+2\times a_y\times s}

v_y=\sqrt{10^2+2\times 10\times 45}

v_y=\sqrt{1000}=31.622\ m/s

Horizontal velocity v_x=u\cos 30=17.32\ m/s

Net velocity Just before hitting =\sqrt{v_x^2+v_y^2}

=\sqrt{(17.32)^2+(31.62)^2}

=\sqrt{1299.82}=36.05\ m/s

                 

7 0
4 years ago
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