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bazaltina [42]
3 years ago
15

Please help me with this question...

Physics
2 answers:
Yakvenalex [24]3 years ago
7 0
1) In first case, both the objects(hand & tea) are in direct contact, so heat will flow by the process of "Conduction" 

2) In this case, objects are not in direct contact, and heat is transferring from one to another by "Convection"

3) Here, there is no contact between the objects (our body & sun), sun light with heat comes through the process of "Radiation"

In short, Your Answers would be:
1) Conduction
2) Convection
3) Radiation

Hope this helps!
Arturiano [62]3 years ago
4 0
The first is conduction. 

This comes as the hot water spills on to your hand through contact.

The second is convection. The hot air rising is due to convection currents being set up.

The third is radiation. This is heat transfer through space.

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A Doppler radar sends a pulse at
Rufina [12.5K]

Answer:

Explanation:

The problem is based on the concept of Doppler's effect of em wave .

Expression for apparent frequency can be given as follows

n = N x (V - v ) / ( V + v )

n is apparent frequency , N is real frequency , V is velocity of light  and v is velocity of cloud.

n = 6 x 10⁹ ( 3 x 10⁸ - 8.52 ) / ( 3 x 10⁸ + 8.52 )

= 6 x 10⁹ ( 3 x 10⁸ ) ( 3 x 10⁸ + 8.52 )⁻¹

= 6 x 10⁹ ( 3 x 10⁸ ) ( 3 x 10⁸)⁻¹ ( 1  + 8.52/3 x 10⁸ )⁻¹

= 6 x 10⁹ ( 1  - 8.52/3 x 10⁸ )

= 6 x 10⁹    - 6 x 10⁹x  8.52/ (3 x 10⁸ )

= 6 x 10⁹  1  - 170 .

So change in frequency = 170 approx.

7 0
3 years ago
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Which forces are shown on a free body diagram?
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Answer:

What is a Free Body Diagram?

The free body diagram helps you understand and solve static and dynamic problem involving forces. It is a diagram including all forces acting on a given object without the other object in the system. You need to first understand all the forces acting on the object and then represent these force by arrows in the direction of the force to be drawn.

Explanation:

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3 years ago
What are the importance of physics
Alexus [3.1K]

Answer:

Physics contributes to the technological infrastructure and provides trained personnel needed to take advantage of scientific advances and discoveries. Physics is an important element in the education of chemists, engineers and computer scientists, as well as practitioners of the other physical and biomedical sciences.

Explanation:

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3 years ago
A shell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. Negl
Triss [41]

Answer:

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g............................. Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

<em>Given: u = 1700 m/s, </em>∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

8 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

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3 years ago
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