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jasenka [17]
2 years ago
10

Pilings are driven into the ground at a building site by dropping a 2250 kg object onto them. What change in gravitational poten

tial energy does the object undergo if it is released from rest 16.0 m above the ground and ends up 1.30 m above the ground ? Change in gravitational potential energy:
Physics
1 answer:
den301095 [7]2 years ago
5 0

Answer: 324.135 kJ

Explanation:

Given

Mass of dropping is m=2250\ kg

The initial height of dropping is h_1=16\ m

The final height of dropping h_2=1.3\ m

Gravitational potential energy is the function of height i.e.

\Rightarrow \text{G.E.}=mgh

Change in Gravitational Energy is

\Rightarrow \Delta \text{G.E.}=mg(h_1-h_2)=2250\times 9.8\times (16-1.3)\\\\\Rightarrow \Delta \text{G.E.}=3,24,135\ J\approx 324.135\ kJ

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At which position would a person on earth see the entire lighted half of the moon?
Gemiola [76]

The best position for the person would be outside, under a clear sky, standing up.  He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.

3 0
3 years ago
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

5 0
3 years ago
A force of 900 N is applied to a 9kg object. How fast will it accelerate?
saul85 [17]

Answer:  15 m/s2

Explanation: I hope this helps or right because I learned this a few months ago

7 0
2 years ago
Read 2 more answers
The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m
algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

E=PE+KE=const.

The potential energy is given by

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

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

E=KE_{max}=\frac{1}{2}mv_{max}^2 (1)

where

m is the mass of the block

v_{max}=1.25 m/s is the maximum speed

We also know that the total energy is

E=0.18 J

Re-arranging eq.(1), we can find the mass:

m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg

b)

The maximum speed in a spring-mass system is also given by

v_{max} =\sqrt{\frac{k}{m}} A

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

v_{max}=1.25 m/s is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m

c)

The frequency in a spring-mass system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz

d)

We can solve this part by using the law of conservation of energy; in fact, we have

E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s

#LearnwithBrainly

3 0
3 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
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