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koban [17]
3 years ago
6

Imagine that someone is sitting down to enjoy a cup of coffee or hot chocolate. Use your experiences to describe how heat flows

from the drink to its surroundings. Be as detailed as possible.
Physics
2 answers:
Reika [66]3 years ago
8 0

The heat flows into my body and I start with a warming feeling going down my throat as it starts there I get this warm feeling and I break out in a coldsweat from consuming a warm drink a cold drink will help cool me down :

Explanation:

If drink warm will make u warm drinking cool will make u cool down faster

il63 [147K]3 years ago
5 0

Answer:

The hot drink heats the air and the cup that’s in direct contact with the drink. The cup heats the person’s hand too. Heat radiates outward from the cup. Heat moves upward when steam is produced.

Explanation:

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AURORKA [14]

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5 0
3 years ago
Why are there only two elements in the first period of the periodic table?(1 point)
tester [92]

Answer:

because each row increases in atomic mass by a specific number, so anything over five is in the second row.

8 0
3 years ago
When you set something down on the ground wht kind of work are your arms doing?
umka2103 [35]
When we set something down on the ground, the kind of work that our arms doing is : negative apex
It's happen whenever we do works that are align with the force of Gravity (to the bottom)

hope this helps
7 0
3 years ago
Read 2 more answers
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
Read 2 more answers
A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0
3 years ago
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