Manipulate the equation "v=d/t" to find the answers to these problems using

Describe how you could measure the thickness of paper with an ordinary ruler

Svetllana [295]

Measure a whole stack (one in which you know the number of sheets), then divide your measurement by the number of sheets in that stack

5 0

3 years ago

when 82.5 calories of heat are given to a metallic rod of mass 150g its temperature raises from 20 degree celcius to 25 degree C

marishachu [46]

A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.

<h3>What is specific heat?</h3>

It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

A metallic rod of mass 150 g (m) absorbs 82.5 cal of heat (Q) and its temperature raises from 20 °C to 25 °C. We can calculate the specific heat (c) of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 82.5 cal / 150 g × (25 °C - 20 °C) = 0.11 cal/g.°C

A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.

Learn more about specific heat here: brainly.com/question/21406849

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8 0

1 year ago

A 1200 kg car accelerates from 13 m/s to 17 m/s. find the change in momentum of the car.

mixas84 [53]

P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s

8 0

1 year ago

Two factors that can be used to evaluate ______ are life expectancy and quality of life.

Olin [163]

Two factors that can be used to evaluate health are life expectancy and quality of life

8 0

2 years ago

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

alexgriva [62]

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

8 0

2 years ago