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krok68 [10]
3 years ago
5

An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec

ond of time? Explain.
a. 2h
b. 4h
c. h
d. h^2
e. 3h
Physics
1 answer:
Viefleur [7K]3 years ago
3 0

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

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the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?
Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

L = length = 10[m]

W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

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density = 1000[kg/m^3]

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h = heigth = 4[cm] = 0.04[m]

P = 1000*9.81*0.04

P = 392.4[Pa]

The force can be easily calculated knowing the relationship between pressure and force:

P = F/A

F = P*A

F = 392.4*1.8

F = 706.32[N]

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Explanation:

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Explanation:

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Please answer ASAP .
Arisa [49]

Answer:

Explanation:

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