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sergejj [24]
3 years ago
9

A ball which is dropped from the top of a building strikes the ground with a speed of 30m/s. Assume air resistance can be ignore

d. The height of the building is approximately
Physics
1 answer:
Vika [28.1K]3 years ago
3 0

Answer:

               h= 45.87 m.

Explanation:

Data given:

Vf= 30m/s ,     and we know that g = 9.8 m/s²    

The ball has an initial velocity of zero      Vi = 0 m/s²    

To Find:

Height of the building = ?

Solution:

       According to 3rd law of the motion;

                                      2aS= Vf² - Vi²                      ( S= h , a=g)

                                     2*9.81*h = (30)² - (0)²

                                      h= 45.87 m.

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An astronaut is walking in space. Which of these would have the greatest speed as observed by the astronaut?
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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
1. The following can be inferred from Newton’s second law of motion except:
lora16 [44]

Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)

2.a)acceleration decreases

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Newton's second law:

Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.

When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.

This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .

7 0
2 years ago
A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C
lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

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v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0
2 years ago
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