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alina1380 [7]
2 years ago
11

what would happen to the temperature on mercury during the day if a sunspot was directly in line with it's surface?

Physics
1 answer:
QveST [7]2 years ago
3 0

Answer:

mind-bogglingly hot

Explanation:

If a sunspot were pointed right at Mercury in this way, the planet would become even more mind-bogglingly hot. Sunspots look dark in our visual spectrum, so we might suspect that they are colder (or at least less hot) than the rest of the Sun; but in fact, they are hotter.

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hot-air balloon is ascending at the rate of 14 m/s and is 84 m above the ground when a package is dropped over the side. (a) How
timurjin [86]

Answer:

a) t = 4.14 s

b) Speed with which it hits the ground = 40.58 m/s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = 84 m,

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) y = ut + gt²/2

84 = 0 + 9.8t²/2

4.9t² = 84

t² = 84/4.9

t = 4.14 s

b) v = u + gt

v = 0 + (9.8 × 4.14)

v = 40.58 m/s

4 0
4 years ago
Read 2 more answers
Consider a semi-spherical bowl machined out of wood (A drawing of a semi-spherical bowl can be found in UNIT- 4 Introduction, so
Mrrafil [7]

Answer:

The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)

Explanation:

The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.

a) Outer semi-sphere:

A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²

b) Inner semi-sphere:

A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²

c) Edge (Ring):

A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²

Therefore, the total surface area of the bowl is given by:

A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)

Changing units to m², as required in the problem, we get:

A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)

5 0
3 years ago
A.) Determine the work done by Zach on the bull.
tresset_1 [31]

Explanation:

The unit of work done is in Joules

Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

    Work done  = Force x distance

The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

 So to solve this problem;

 multiply the force applied by Zack and distance through which the bull was pulled.

8 0
3 years ago
An autotransformer is used to reduce the voltage of a 100-kilovolt amp, 480-volt secondary of an isolated type transformer, to s
tigry1 [53]

Answer:

42KVA

Explanation:

Given data

High Voltage (HV)= 480V

Low Voltage (LV)= 277V

Fo find

Size of transformer=?

Solution

To find the size of transformer here we use the co-ratio.The Co-ratio is given as:

Co-Ratio= (HV - LV)/HV

where

HV is High Voltage

LV is Low Voltage

Now put the values we get

Co- Ratio=(480-277)/480=.42

So the size of transformer is 42KVA

5 0
3 years ago
What force does a trampoline have to apply to a gymnast to accelerate her straight up at ? Note that the answer is independent o
Andrew [12]

Answer: Force applied by trampoline = 778.5 N

<em>Note: The question is incomplete.</em>

<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary. </em>

Explanation:

The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.

= (m * a) + (m * g)

= m ( a + g)

= 45 kg ( 7.50 *  9.80) m/s²

Force applied by trampoline = 778.5 N

5 0
3 years ago
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