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Licemer1 [7]
3 years ago
10

Can someone help me with these physics questions?

Physics
1 answer:
wariber [46]3 years ago
6 0

hi! Glad to help =)

1. would be c

2 would be d

3 would be A

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What is the kinetic energy of a car of 1500 kg moving 25 m/s?
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Is this a multi choice?
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3 years ago
Put the following objects in order by their
maksim [4K]

Answer:

I Think its ABC

Explanation:

No air is the lightest

Helium is lighter than regular air that's why it goes up so then the regular air would be heaviest.

4 0
3 years ago
A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when
ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, V_T = 60 \ m/s

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

F_D = kV_T^2

Where;

k is a constant

k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}

When the new drag force is half of the original drag force;

F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}=  \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}=  0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s

Therefore, the speed of the resistive force is 42.426 m/s

8 0
3 years ago
What is Planck’s law?
Lyrx [107]

Answer:

The wavelength of the emitted radiation is inversely proportional to its frequency, or λ = c/ν. The value of Planck's constant is defined as 6.62607015 × 10−34 joule∙second.

Explanation:

Planck's quantum theory. According to Planck's quantum theory, Different atoms and molecules can emit or absorb energy in discrete quantities only. The smallest amount of energy that can be emitted or absorbed in the form of electromagnetic radiation is known as quantum.

Hope this helps!

3 0
3 years ago
In the diagram, q1 = +6.39*10^_9 C and
ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

q_1=6.39\times 10^{-9} \ C

q_2=3.22\times 10^{-9} \ C

AP=(0.150+0.250)

      =0.40 \ m

BP=0.25 \ m

Now,

At point P, the electric potential will be:

⇒ V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}

By putting values, we get

⇒     =9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]

⇒     =259.695 \ Volt

4 0
3 years ago
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