Answer:

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♨ Question :
- A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?
♨ 
☄ Given :
- Initial velocity ( u ) = 0
- Final velocity ( v ) = 60.0 m/s
- Time ( t ) = 7.50 s
☄ To find :
✒ We know ,

Substitute the values and solve for a.
➛ 
➛ 
➛ 
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✑ Additional Info :
- When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
- When a moving object comes in rest , in the case , final velocity ( v ) = 0
- If the object is moving with uniform velocity , in the case , u = v.
- If any object is thrown vertically upwards in the case , a = -g
- When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.
Hope I helped!
Have a wonderful time ツ
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M₁ = mass of planet #1
M₂ = mass of planet #2
M = total mass
R₁ = radius of planet #1
R₂ = radius of planet #2
d₁ = initial distance between planet centers
d₂ = final distance between planet centers
a = semimajor axis of plunge orbit
v₁ = relative speed of approach at distance d₁
v₂ = relative speed of approach at distance d₂
M₁ = M₂ = 1.8986e27 kilograms
M = M₁ + M₂ = 3.7972e27 kg
G = 6.6742e-11 m³ kg⁻¹ sec⁻²
GM = 2.5343e17 m³ sec⁻²
d₁ = 1.4e11 meters
a = d₁/2 = 7e10 meters
R₁ = R₂ = 7.1492e7 meters
d₂ = R₁ + R₂ = 1.42984e8 meters
v₁ = 0
v₂ = √[GM(2/d₂−1/a)]
<span>
v₂ = 59508.4 m/s </span>
<span>
The time to fall is 1337.7 days
.</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly.
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance,
= 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
Answer:
Deoxyribonucleic acid (DNA) is a nucleic acid that contains the genetic instructions for the development and function of living things. All known cellular life and some viruses contain DNA. The main role of DNA in the cell is the long-term storage of information.
link: https://www.sciencedaily.com/terms/dna.htm
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The force = 0.3 * 9.8 = 2.94
x = 0.13
k = 2.94/0.13
k = 22.6 N/m