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DENIUS [597]
2 years ago
5

A gardener pushes a wheelbarrow around the edges of a lawn. When she has finished, she is 5 m from her starting point. She has t

o pull upwards with a force of 300 N to lift the legs of the wheelbarrow off the ground. She pushes the wheelbarrow along the ground with a force of 100 N. Calculate how much work the gardener does in pushing the wheelbarrow around the lawn.
Physics
1 answer:
zvonat [6]2 years ago
3 0

Answer:

Work done by the gardner is 500 J

Explanation:

As we know that the gardner apply force perpendicular upward by magnitude 300 N and along the floor horizontal force is 100 N

so we have

F = 100 \hat i + 300 \hat j

now the displacement of the gardner along the floor is

d = 5\hat i

now work done is given as

W = F. d

so we have

W = (100 \hat i + 300 \hat j). (5\hat i)

W = 500 J

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How many electrons make up a charge of - 30.0 μc?
Ymorist [56]

Each electron has a charge of

q=-1.6 \cdot 10^{-19}C

In this problem, the total charge is

Q=-30.0 \mu C=-30.0 \cdot 10^{-6} C

Therefore, the number of electrons contained in this total charge will be given by the total charge divided by the charge of a single electron:

N=\frac{Q}{q}=\frac{-30.0 \cdot 10^{-6} C}{-1.6 \cdot 10^{-19} C}=1.88 \cdot 10^{14}

3 0
3 years ago
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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
What is meant by sundial???​
Fofino [41]

Answer:

Sundial is an instrument showing the time by the shadow of a pointer cast by the sun on to a plate marked with the hours of the day.

4 0
2 years ago
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A toy rocket is fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the time it wil
miskamm [114]

Answer: 16.3 seconds

Explanation: Given that the

Initial velocity U = 80 ft/s

Let's first calculate the maximum height reached by using third equation of motion.

V^2 = U^2 - 2gH

Where V = final velocity and H = maximum height.

Since the toy is moving against the gravity, g will be negative.

At maximum height, V = 0

0 = 80^2 - 2 × 9.81 × H

6400 = 19.62H

H = 6400/19.62

H = 326.2

Let's us second equation of motion to find time.

H = Ut - 1/2gt^2

Let assume that the ball is dropped from the maximum height. Then,

U = 0. The equation will be reduced to

H = 1/2gt^2

326.2 = 1/2 × 9.81 × t^2

326.2 = 4.905t^2

t^2 = 326.2/4.905

t = sqrt( 66.5 )

t = 8.15 seconds

The time it will take for the rocket to return to ground level will be 2t.

That is, 2 × 8.15 = 16.3 seconds

8 0
3 years ago
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d1i1m1o1n [39]
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
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T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
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3 years ago
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