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makvit [3.9K]
3 years ago
9

science numerical problem .Hari drag a load a of 60 kg along a distance of 12 m . what amount of work does he do? as 0 mention t

he type of work​
Physics
1 answer:
Komok [63]3 years ago
6 0

We don't know, and we don't have enough information to calculate it.

The weight of the 60kg load is (m g) = 588 Newtons.

IF Hari wanted to<em> lift </em>the load 12m <em>straight up</em>, he would have to do

(Force x distance) = (588 N) x (12 m) =  7,056 Joules of work.

But to drag it, he has to provide enough force to balance out the force of friction, and we don't know how much that is. It depends on the weight of the load, the shape of the load, the smoothness of the part of the load that sits on the ground, and the smoothness of the ground.  But the only thing we know is the weight of the load.

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Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0
3 years ago
QUESTION 1 The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 968 Hz?​
Anton [14]

Answer:

Explanation:

340 m/s / 968 cyc/s = 0.3512396... ≈ 35.1 cm

7 0
3 years ago
when a constant force acts upon an object the acceleration of the object varies inversely with its mass. when a certain constant
solong [7]

Explanation:

When a constant force acts upon an object the acceleration of the object varies inversely with its mass.

a\propto \dfrac{1}{m}

or

\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}

If m₁ = 21 kg, a₁ = 3 m/s², m₂ = 9 kg

We need to find a₂

So,

a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{21\times 3}{9}\\\\a_2=7\ m/s^2

So, if mass is 9 kg, its acceleration is 7 m/s².

8 0
2 years ago
QUESTION 1
lisabon 2012 [21]

Answer:

2 * 10^5 pa

Explanation:

Pressure = Force / Area

Each thigh bone has a cross sectional area of 10cm²

Both thigh bones :

2 * 10cm² = 20cm²

To m² : 20 * (0.01)²

20 * 0.0001 m² = 0.002 m²

Force = mass * acceleration due to gravity(g)

g = 10m/s² ;

Force = 40 * 10 = 400N

Pressure = 400 N / 0.002 m²

Pressure = 200,000 N/m² = 2 * 10^5 pascal

8 0
2 years ago
on an unknown temperature scale, the freezing point of water is -15°U and the boiling point is +60°U. develop a linear conversio
shutvik [7]

Answer:

Since this is a linear equation

y = m x + b     or

U = m F + b     is a linear equation

when ΔF = (212 - 32) = 180

and ΔU = (60 - (-15)) = 75

m = 75 / 180 = 2.4 if converting F to U and a = .417

U = .417 F + b

If F = 32 then U = -15 and

-15 = .417 * 32 + b

b = -15 - 13.3 = -28.3 and our equation becomes

U = .417 F - 28.3

Check: let F = 212

U = .417 * 212 - 28.3 = 60          as it should

5 0
1 year ago
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