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Alecsey [184]
2 years ago
12

During a solar eclipse our tiny moon totally blocks the sun because of .....

Physics
1 answer:
Vaselesa [24]2 years ago
4 0

During a total solar eclipse, the moon passes between Earth and the sun. This completely blocks out the sun’s light. However, the moon is about 400 times smaller than the sun. How can it block all of that light?

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Where does energy originate ina food web
ValentinkaMS [17]

\huge{\underline{\underline{\red{Answer}}}}

In a food web the energy is originated from the SUN.

4 0
2 years ago
Read 2 more answers
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward
Anni [7]

Answer:

m=18000kg

Explanation:

From the question we are told that:

Crane Length l=20m

Crane Mass m_a=3000kg

Arm extension at lifting end l_l=15m

Arm extension at counter weight end l_c=5m

Load M_l=5000kg

Generally the equation for Torque Balance is mathematically given by

T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0

mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0

m=18000kg

7 0
3 years ago
In the us we use the us customary units true or false
sergiy2304 [10]

Answer:

true

Explanation:

i believe that's the answer hope that helps

6 0
3 years ago
A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is m
konstantin123 [22]

Answer:

The torque on the loop is 2.4 \times 10^{-2} Nm

Explanation:

Given:

Current I = 1.6 A

Magnetic field B = 0.30 T

Area of loop A = 0.14 m^{2}

Angle between magnetic field and area vector \theta = 21°

Form the formula of  torque in case of magnetic field,

 г = MB \sin \theta

Where M = magnetic moment

  M = IA

 г = IAB \sin 21

 г = 1.6 \times 0.30 \times 0.14 \times 0.3583

 г =2.4 \times 10^{-2} Nm

Therefore, the torque on the loop is 2.4 \times 10^{-2} Nm

7 0
3 years ago
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