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zmey [24]
2 years ago
5

Please answer of this question​

Physics
1 answer:
Bogdan [553]2 years ago
8 0

Answer:

\frac{\pi }{15}m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = 2\pi r(\frac{60}{360} )

= 2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m

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200 what factors do Pressure depend?<br>1 What is a<br>thematical expression do you u​
Varvara68 [4.7K]

Answer:

pressure of solid depends on:

1/magnitude of the force

2/contact area

pressure in solid depends on:

1/depth inside the fluid.

2/density of the fluid.

3/Acceleration due to gravity.

Explanation:

1/magnitude of the force:the pressure is directly proportional to the magnitude of the force(thrust). The larger the force, the higher is the magnitude of pressure on the surface.

2/Contact area: the pressure is inversely proportional to the surface area of contact. The larger the contact area, lower is the magnitude of the pressure.

6 0
2 years ago
I need these for test corrections that are due tomorrow please :)
Ivahew [28]
That would be position Y, as the northern hemisphere is tilted away from the sun.
6 0
3 years ago
Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second
REY [17]

Answer:

E = 75 J

Explanation:

First, we will calculate the total power consumed by the five lamps:

Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W

Now, the energy supply can be calculated as follows:

E = Pt

where,

E = Energy = ?

t = time = 5 s

Therefore,

E = (15 W)(5 s)

<u>E = 75 J</u>

8 0
2 years ago
The distance between two charged objects is doubled. What happens to the electrostatic force between the two?a)It will double.b)
zzz [600]

Answer:

d) It will be cut to a fourth of the original force.

Explanation:

The magnitude of the electrostatic force between the charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, the initial distance is doubled, so

r' = 2r

Therefore, the new electrostatic force will be

F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F

So, the force will be cut to 1/4 of the original value.

5 0
3 years ago
On an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. The children push off of
sasho [114]

The velocity of the girl is  -4.8 m/s.

Using the principle of conservation of linear momentum, The total momentum of  bodies before and after collision is constant. Since the two objects are stationary, the initial momentum of each body is zero.

Thus;

0 = (80 × 3) + (50 × v)

0 = 240 + 50 v

-240 = 50 v

v = -240/50

v = -4.8 m/s

Note that the negative sign shows that the velocity of the girl is in opposite direction that that of the girl.

Learn more about momentum: brainly.com/question/904448

5 0
3 years ago
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