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3 years ago

6

If a person tries to lift a heavy box for 5 seconds and can't make it budge,the work done on the box is less than zero

1 answer:

blondinia [14]3 years ago

3 0

Question not making a sence, Clarify what you wana ask

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In a shipping yard, a crane operator attaches a cable to a 1,400 kg shipping container and then uses the crane to lift the conta

Molodets [167]

Answer:

Explanation:

Given

mass of crane $m=1400\ kg$

distance moved $d=40\ m$

Since it is moving with a constant velocity therefore net force on it is zero

Tension force=weight

T=mg

Work done by Tension T is

$W_T=T\cdot d$

$W_T=1400\times 9.8\cdot 40$

$W_T=548.8\ KJ$

Work done by Gravity will be equal in magnitude but opposite in sign and can be obtained by work energy theorem which states that change in kinetic energy of object is equal to work done by all the forces

$W_T+W_g=0$

$W_g=-548.8\ KJ$

4 0

3 years ago

An electric field of 3x10^16 n/c is needed to create a spark in the air. If the charged particles in the field are separated by

meriva

The answer of this problem is positive 3

6 0

3 years ago

The rate at which the earth's surface loses heat and keeps earth temperature at which living things can survive

agasfer [191]

<span>I believe it's insulation.</span>

5 0

3 years ago

This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi

Rufina [12.5K]

Answer:

N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

N - w₁ -w₂ =

N = m₁ g + m₂ g

N = g (m₁ + m₂)

let's calculate

N = 9.8 (0.760 + 1.630)

N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0

3 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

Llana [10]

Answer:

a) $C = 40.138\,pF$ , b) $q = 16.056\,nC$ , c) $U = 3.212\,\mu J$

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

$C = K\cdot \epsilon_{o}\cdot \frac{A}{d}$

Where:

$K$ - Dielectric constant.

$\epsilon_{o}$ - Vaccum permitivity.

$A$ - Plate area.

$d$ - Distance between plates.

Hence, the capacitance of the system is:

$C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)$

$C = 4.014\cdot 10^{-11}\,F$

$C = 40.138\,pF$

b) The charge can be found by using the definition of capacitance:

$q = C\cdot V_{batt}$

$q = (4.014\times 10^{-11}\,F)\cdot (400\,V)$

$q = 1.606\times 10^{-8}\,C$

$q = 16.056\,nC$

c) The energy stored in the charged capacitor is:

$U=\frac{1}{2}\cdot Q\cdot V_{batt}$

$U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)$

$U = 3.212\times 10^{-6}\,J$

$U = 3.212\,\mu J$

3 0

3 years ago

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