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2 years ago

Circle the author's purpose for writing this passage

Physics

2 answers:

Naily [24]2 years ago

5 0

The author's purpose is to persuade

Gnesinka [82]2 years ago

5 0

The authors purpose would be that he would like to see more people roller blading because its a good form of exercise and can really boost your health up

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How does energy from the Sun travel to Earth?

bezimeni [28]

Answer:

the energy from the sun travel to earth the answer is A .through the radiation

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2 years ago

Read 2 more answers

A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

$\vec{F} = q \vec{E}$

$11\hat{i} = -5 \times 10^{-6} \vec{E}$

$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

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2 years ago

Answer this plsss<br> A mobile phone operates at 900 MHz. <br> What wavelength does it use?

Aleks [24]

Answer:

The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.

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2 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

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3 years ago

Choose all the answers that apply.

Serhud [2]

<span>According to Newton's first law of motion:

-- objects at rest will remain at rest unless acted upon by an outside force

-- objects in motion will remain in motion unless acted upon by an outside force

</span>

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2 years ago