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storchak [24]
2 years ago
12

A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds. Wha

t is the acceleration? 0-13 m/s2 0-6 m/s2 0.6 m/s2 O 13 m/s2​
Physics
1 answer:
olga55 [171]2 years ago
3 0

Answer:

6m/s²

Explanation:

Given parameters:

Initial velocity  = 35m/s

Final velocity  = 65m/s

Time taken  = 5s

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time taken;

   Acceleration  = \frac{Final velocity  - Initial velocity }{time}  

So;

 Acceleration   = \frac{65  - 35}{5}    = 6m/s²

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Varg sees a spring that has a spring constant of 4 N/m that is stretched 5 m. He stretches the spring an additional 5 m. Conside
AysviL [449]

Answer:

Elastic potential energy, E = 200 J

Explanation:

It is given that,

Spring constant, K = 4 N/m

initial stretching in the spring, x = 5 m

Finally, it is stretched an additional 5 m i.e. x' = 5 m        

Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :

E=\dfrac{1}{2}k(x+x')^2

E=\dfrac{1}{2}\times 4\times (10)^2

E = 200 J

So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.

8 0
3 years ago
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If a vector that is 3cm long represents 30 km/h, what velocity does a 5 cm long vector which is drawn using the same scale repre
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Answer:

False knowww yess ofcure

8 0
3 years ago
An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:
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..... It would possibly she eenejjsjejeej 1.4
4 0
3 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL?<br> Show work!
Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

 657

×  32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

   ↳            21024 g/mL

3 0
3 years ago
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