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3 years ago

How do you make magnetic and static forces strong or week?

1 answer:

6 0

You make them strong for example if you rub a balloon in your hair the longer you do it the longer it sticks because of static

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An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago

What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.

Stolb23 [73]

Answer:

$Momentum(p) = 20kgm/s$

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

$Momentum(p) = Mass(m) * Velocity(v)$

$p = mv$

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

$Momentum(p) = 10kg * 2m/s$

$Momentum(p) = 10 * 2 * kg * m/s$

$Momentum(p) = 20kgm/s$

Hence, the momentum of the man is $20kgm/s$

5 0

3 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff wit

pantera1 [17]

Answer:

height of a cliff is 45.86 m

Explanation:

u=8 m/s

t=2.35 s

the height of the cliff is

$S=ut+\frac{1}{2} at^2\\S=8\times2.35+\frac{1}{2} \times 9.8 \times 2.35^2\\S=45.86 m$

7 0

3 years ago

Conduction can only happen when objects are in:

serious [3.7K]

Answer:

When it’s gravitational

Explanation:

I’m pretty sure that’s the answer

5 0

3 years ago

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