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hammer [34]
3 years ago
6

Please help me on questions 3 and 6, thank you! :D I'll give brainliest!

Physics
2 answers:
KIM [24]3 years ago
6 0

3. To increase gravitational pull to the earth, the more mass an object has, the stronger the gravitational pull, so increase mass.

6. While sound energy depends on a medium, radiant energy uses it's own sources to stay independent from other mediums.

Anarel [89]3 years ago
4 0

Answer:

3. if you increase your mass you also increase the gravitational pull

6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.

Explanation:

i hope this helps-

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Categorize the particles according to whether they are found in the nucleus or the electron cloud.
inna [77]
1. neutral particles (neutrons) are in the nucleus
2. nucleus is in the nucleus
3. electron cloud is in the electron cloud
4. positively charged particles (protons) are in the nucleus
5. negatively charged particles (electrons) are in the electron cloud
4 0
3 years ago
Read 2 more answers
1) A car is moving to the right with the constant speed of 1.2 m/s. If the car starts
Marina CMI [18]
(100, 108)
Due to
1.2x90=108

100, 108
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3 years ago
At 20∘C20 ∘ C, the hole in an aluminum ring is 2.500 cm in diameter. You need to slip this ring over a steel shaft that has a ro
Feliz [49]

Answer:

238.75⁰C .

Explanation:

coefficient of linear thermal expansion of aluminum and steel is  23 x 10⁻⁶ K⁻¹ and 12 x 10⁻⁶ K⁻¹ respectively .

Rise in temperature be Δ t .

Formula for linear expansion due to heat is as follows

l = l₀ ( 1 + α x Δt )

l is expanded length , l₀ is initial length , α is coefficient of linear expansion and  Δt is increase in temperature .

For aluminum

l = 2.5 ( 1 +  23 x 10⁻⁶ Δt )

For steel

l = 2.506 ( 1 + 12 x 10⁻⁶ Δt )

Given ,

2.5 ( 1 +  23 x 10⁻⁶ Δt ) = 2.506 ( 1 + 12 x 10⁻⁶ Δt )

1 +  23 x 10⁻⁶ Δt = 1.0024 ( 1 + 12 x 10⁻⁶ Δt )

1 +  23 x 10⁻⁶ Δt = 1.0024   + 12.0288 x 10⁻⁶ Δt

10.9712 x 10⁻⁶ Δt = .0024

Δt = 218.75

Initial temperature = 20⁰C

final temperature = 218.75 + 20 = 238.75⁰C .

8 0
3 years ago
A 15,000 kg rocket traveling at +230 m/s turns on its engines. Over a 6.0 s period it burns 1,000 kg of fuel. An observer on the
lesya692 [45]

Answer:

a) v = 312.791\,\frac{m}{s}, b) a = 13.333\,\frac{m}{s^{2}}

Explanation:

The problem is asking the rocket velocity and acceleration at t = 6 s.

a) The general equation of the rocket is:

v=v_{o} -v_{ex}\cdot \ln \frac{m}{m_{o}}

v = 230\,\frac{m}{s}-(1200\,\frac{m}{s} )\cdot \ln \frac{14000\,kg}{15000\,kg}

v = 312.791\,\frac{m}{s}

b) The acceleration experimented by the rocket is:

a = \frac{v_{ex}}{m_{o}}\cdot \frac{dm}{dt}

a = \frac{1200\,\frac{m}{s} }{15000\,kg}\cdot \frac{1000\,kg}{6\,s}

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3 0
4 years ago
Read 2 more answers
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
Serga [27]

Answer:

115 m/s, 414 km/hr

Explanation:

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∑F = ma

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D = mg

Drag force is defined as:

D = ½ ρ v² C A

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v is the velocity,

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Substituting and solving for v:

½ ρ v² C A = mg

v² = 2mg / (ρCA)

v = √(2mg / (ρCA))

We're given values for m and A, and we know the value of g.  We need to look up ρ and C.

Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.

For a skydiver falling headfirst, C ≈ 0.7.

Substituting all values:

v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))

v = 115 m/s

v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)

v = 414 km/hr

4 0
3 years ago
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