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3 years ago

When the car was stopped by the tree, its change in velocity during the collision was-6 meters/second. This change in

Physics

1 answer:

HACTEHA [7]3 years ago

3 0

Answer: -3 meters/second^2.

Explanation:

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How does heat affect the thermal energy of an object that is colder than the qir

podryga [215]

Answer:

the correct answer is B

Explanation:

thermal energy makes heat increase.

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3 years ago

Which type of electromagnetic radiation cannot be focused?

Irina18 [472]

The answer is A. Hope this helps. :)

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3 years ago

Read 2 more answers

What is the value of the charge that experiences a force of 2.4 Ã— 10â€“3 n in an electric field of 6.8 Ã— 10â€“5 n/c?

tia_tia [17]

Answer:

The value of charge is 35.3 C

Explanation:

Given:

Force $F = 2.4 \times 10^{-3}$ N

Electric field $E = 6.8 \times 10^{-5} \frac{N}{C}$

According coulomb's law,

$F = qE$

Where $q =$ charge that experience a force,

For finding the value of charge,

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

$q = \frac{2.4 \times 10^{-3} }{6.8 \times 10^{-5} }$

$q = 35.3$ C

Therefore, the value of charge is 35.3 C

6 0

3 years ago

At what net rate does heat radiate from a 300-m^2 black roof on a night when the roof's temperature is 33.0°C and the surroundin

Fynjy0 [20]

Answer:

24445.85 J/s

Explanation:

Area, A = 300 m^2

T = 33° C = 33 + 273 = 306 k

To = 18° C = 18 + 273 = 291 k

emissivity, e = 0.9

Use the Stefan's Boltzman law

$E = \sigma \times e \times A\times\left ( T^4 -T_{0}^{4}\right )$

Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.

The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4

By substituting the values

$E = 5.64 \times 10^{-8}\times 0.9 \times 300 \times (306^{4}-291^{4})$

E = 24445.85 J/s

7 0

4 years ago

The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the

ycow [4]

$F = 5.93×10^{13}\:\text{N}$

Explanation:

Given:

$m_1= 2×10^{16}\:\text{kg}$

$m_2= 4×10^{22}\:\text{kg}$

$r = 30000\:\text{km} = 3×10^7\:\text{m}$

Using Newton's universal law of gravitation, we can write

$F = G\dfrac{m_1m_2}{r^2}$

$\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}$

$\:\:\:\:= 5.93×10^{13}\:\text{N}$

3 0

3 years ago