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n200080 [17]
3 years ago
14

As the time required to run up the stairs increases, the power developed by that person

Physics
1 answer:
weeeeeb [17]3 years ago
8 0
Yes. Power will decrease.

'cause Power = Work / time
So, power is indirectly proportional to time so, when one increases other would decrease

Hope this helps!
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The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?
Marat540 [252]
Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is. 
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What is defined by force per unit area
sleet_krkn [62]

Explanation:

<em>For</em><em>ce</em><em> </em><em>per</em><em> </em><em>uni</em><em>t</em><em> </em><em>area</em><em> </em><em>is</em><em> </em><em>call</em><em>ed</em><em> </em><em>Pres</em><em>sure</em><em>.</em>

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3 years ago
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A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above
gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

F_s = \mu N

where \mu is the coefficient of friction and N is the normal force.

When the student pulls on the block with force F_p at an angle \theta, the normal force on the block becomes

N  = Mg- F_psin(\theta)

and hence the frictional force becomes

F_s = \mu (Mg- F_psin(\theta)).

Now, as we increase \theta, sin(\theta) increases which as a result decreases the normal force Mg- F_psin(\theta), which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em>W=mg<em> is independent of the external forces on the block. </em>

6 0
4 years ago
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Julia can swim at 3.5 km/h in still water. She attempts to head straight north
raketka [301]

1) 3.7 km/hr  N19°W

2) she must aim at N20°E

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3 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
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