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BabaBlast [244]
3 years ago
9

True or false Compared to water, most metals have low specific heats.

Physics
2 answers:
masha68 [24]3 years ago
5 0
The answer would be true
elena-14-01-66 [18.8K]3 years ago
4 0
Hey there,

Answer:

True

Hope this helps :D

<em>~Top♥</em>
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How might you use a rope and pulley to move a wagon up a ramp?
sleet_krkn [62]
You could attach the pulley to a secure object on the top of the ramp, and crank the pulley to bring the wagon up said ramp into a loading bay perhaps, or a track. 

Hope I helped. 
7 0
3 years ago
If you apply a force of 100 N to the level, how much force is applied to lift the crate?
Kaylis [27]
I believe that the answer is B. 133 N
4 0
3 years ago
A runner begins a race from the starting line and accelerates to a speed of 8.9 m/s. If it takes the runner 3 seconds to reach h
stellarik [79]

Answer:

i believe its 26.7

Explanation:

if the runner goes 8.9 m/s each second while accelerating for 3 seconds to reach top speed, the top speed would be 26.7 m/s

4 0
2 years ago
Read 2 more answers
A child is sliding down a slide at the playgound. is mechanicalenergy conserved
Flauer [41]

No. Mechanical energy is not conserved.  There's quite a bit of friction on the slide.  So some of the potential energy is lost to heat on the way down, and the child arrives at the bottom with hot pants and less kinetic energy than you might expect.

5 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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