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Natali [406]
3 years ago
13

5. He slept for eight hours. This sentence has

Physics
2 answers:
Galina-37 [17]3 years ago
7 0

Answer:

an object (it's English

Alisiya [41]3 years ago
6 0
An object ( it’s english )
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A bus accelerates 25 m/s" and the bus has a mass of 150kg. What is the force on the bus?
o-na [289]

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150 I would believe that it is the correct answer

4 0
3 years ago
Young's double slit experiment is one of the classic tests for the wave nature of light. In an experiment using red light (λ = 6
Marianna [84]

Answer:

The separation distance between the slits is 16710.32 nm.

Explanation:

Given that,

Wavelength = 641 nm

Angle =4.4°

(a). We need to calculate the separation distance between the slits

Using formula of young's double slit

d\sin\theta=m\lambda

d=\dfrac{m\lambda}{\sin\theta}

Where, d = the separation distance between the slits

m = number of order

\lambda =wavelength

Put the value into the formula

d=\dfrac{2\times641\times10^{-9}}{\sin4.4}

d=0.00001671032\ m

d=16710.32\ nm

Hence, The separation distance between the slits is 16710.32 nm.

3 0
3 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
In order for electric current to flow, an electric circuit must be _____. closed open
krok68 [10]
The circuit has to be closed so current flows.
6 0
3 years ago
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