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Ivahew [28]
2 years ago
7

(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en

ergy did this object gain?
Physics
1 answer:
V125BC [204]2 years ago
7 0

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

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If a 2 kg spring is compressed by exerting a force of 10 newtons a distance of 50 cm how much work is done?
Elina [12.6K]
Convert cm into meters,
50/100=0.5m

work done= 10x0.5
=5J

we dont consider the weight of the spring as it acts downwards.
5 0
3 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
a ladybug sits at the outer edge of a merry-go-round and a gentleman bug sits halfway between her and the axis of rotation. The
forsale [732]
The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)
6 0
2 years ago
The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa
kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0
2 years ago
How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?
Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

d =  \frac{ \:Δv}{e}

=  \frac{1.5v}{100v/m}

= 1.5 \times 10 {}^{ - 2} m

learn more about potential difference from here: brainly.com/question/28166044

#SPJ4

6 0
9 months ago
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