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klasskru [66]
2 years ago
5

URGENT!!!!

Physics
1 answer:
Nastasia [14]2 years ago
3 0

Answer:

8 cm

Ladder is safer than to walk on it

Explanation:

Streched = Final length - initial length

= 66 - 58

= 8 cm

When your are walking along frozen ice pond your weight is not distributed .It's acting on directly on the ice pond because of contact Area is too small between pond and you. There for According to pressure equation ,

P = F/A

When area is less pressure is less.

If use ladder instead of going by foot. Contact area will be high between pond and ladder. There for pressure is reduced .

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Isabella drops a pen off her balcony by accident while celebrating the successful completion of a physics problem. assuming air
expeople1 [14]
U = 0, initial vertical velocity

Neglect air resistance, and g = 9.8 m/s².

The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s

Answer:  2.0 s
8 0
2 years ago
A cyclist moving towards right with an acceleration of 4m/s² at t = 0 he has travelled 5 m moving towards the right at 15 m/s wh
ira [324]

Answer:

x=2t^2+15t+5

Explanation:

x=\frac{1}{2}at^2+v_0t+x_0

4 0
2 years ago
As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
Andreyy89

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

4 0
2 years ago
Read 2 more answers
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put
olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

6 0
2 years ago
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initia
prisoha [69]

Answer:

The water will flow at a speed of 3,884 m/s

Explanation:

Torricelli's equation

v = \sqrt{2gh}

*v = liquid velocity at the exit of the hole

g = gravity acceleration

h = distance from the surface of the liquid to the center of the hole.

v = \sqrt{2*9,8m/s^2*0,77m} = 3,884 m/s

6 0
2 years ago
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