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Brut [27]
3 years ago
6

All objects at any temperature emit radiant energy true or false

Physics
2 answers:
aleksandrvk [35]3 years ago
8 0
It is true.from the sun to ice
kirill [66]3 years ago
4 0
The answer is true.
All object at any temperature emit radiant energy.
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The West the cost
poizon [28]

Answer:

Sck my p3nis

Explanation:

if you do so, then your mom will have coronavirus.

4 0
3 years ago
A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0
3 years ago
Define an element and give 5 examples of elements that are important to life .
fenix001 [56]

Answer:

A pure substance consisting only of atoms with the same number of protons in their nuclei-these appear on the periodic table

Oxygen

Hydrogen

Carbon

Sulfur

Phosphate

Nitrogen

Magnesium

Calcium

Potassium

Chlorine

(I know that these are more examples than needed, but you can use any)

8 0
3 years ago
Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the
Dafna11 [192]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Heat capacity of body 1 :

\qquad \sf  \dashrightarrow \:m_1s_1

Heat capacity of body 2 :

\qquad \sf  \dashrightarrow \:m_2s_2

it's given that, the the head capacities of both the objects are equal. I.e

\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2

\qquad \sf  \dashrightarrow \:m_1 =  \dfrac{m_2s_2}{s_1}

Now, consider specific heat of composite body be s'

According to given relation :

\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }

[ since, m_2s_2 = m_1s_1 ]

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{  \cancel{m_2}(\frac{s_2}{s_1} + 1)}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 s_2}{  (\frac{s_2 + s_1}{s_1} )}

\qquad \sf  \dashrightarrow \: s' =  \dfrac{2s_1s_2}{s_1 + s_2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

6 0
3 years ago
Read 2 more answers
if you use the compound pulley, you will need to pull twice the distance but with less force. the force you need is equal to one
algol13

Answer:

F= 25/2 = 12.5N

Explanation:

When you use a compound pulley the force required depends on the mechanical advantage of the compound pulley. This is known as rate of loss of distance or the ratio of the force to the load.

M.A = Effort distance /Load distance. OR M.A = Load/Effort

6 0
4 years ago
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