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3 years ago

The amount of heat needed to increase the temperature of a 1-kilogram substance by 1°C is known as the

2 answers:

5 0

The amount of heat needed to increase the temperature of a 1-kilogram substance by 1°C is known as the specific heat of the substance.

the formula for specific heat of a substance is given as

c = Q/(m ΔT)

where Q = Heat required to change the temperature by 1°C

m = mass of the substance

ΔT = change in temperature.

the units of specific heat is given as Joules/(kilogram °C)

erastova [34]3 years ago

3 0

Specific heat.

"Different substances gain heat energy at different rates. A 1-kilogram substance's specific heat is a measure of how much heat must be added to it in order for its temperature to increase by 1°C."

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A charged particle of mass 6.00 mg moves with a speed of 4.00 km/s in a direction that makes an angle of 37o above the positive

AysviL [449]

Answer:

Explanation:

mass of charged particle m = 6 x 10⁻³ kg .

speed of particle v = 4 x 10³ m /s

speed of particle perpendicular to magnetic field = v sin37

= 4 x 10³ sin37

= 2.41 x 10³ m / s

Force on charged particle

= B q v , B is magnetic field , q is charge on particle and v is velocity perpendicular to B

Force = ma

= 6 x 10⁻³ x 8

= 48 x 10⁻³

Force = Bqv

48 x 10⁻³ = 5 x 10⁻³ q x 2.41 x 10³

q = 48 x 10⁻³ / (5 x 2.41)

= 3.98 x 10⁻³C.

5 0

4 years ago

Read 2 more answers

What distance is a book from the floor if the book contains 196 joules if potential energy and has a mass of 5kg

vampirchik [111]

196 Joules = (5Kg)(9.8 m/s2)(h)

(196 Joules /(5Kg)(9.8 m/s2) = h

196 Joules / 49 = h

h = 4 meters

Hope this helped!!!!

6 0

3 years ago

A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of

inna [77]

Answer:

The magnitude of the average frictional force on the block is 2 N.

Explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :

$v^2-u^2=2ad$

$-u^2=2ad$

$a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2$

The frictional force on the block is given by the formula as :

F = ma

$F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N$

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

8 0

4 years ago

You are loading a toy dart gun, which has two settings, the more powerful with the spring compressed twice as far as the lower s

Amiraneli [1.4K]

Answer:

option A

Explanation:

given,

compression for lower setting = x

work done to compress in lower setting = 5 J

compression in higher setting, x' = 2 x

work done in higher setting = ?

Work done in compression of spring at lower setting

$W = \dfrac{1}{2}kx^2$

$5 = \dfrac{1}{2}kx^2$ ............(1)

Work done in compression of spring at higher setting

$W' = \dfrac{1}{2}kx'^2$

$W'= \dfrac{1}{2}k(2x)^2$

$W'= 4\times \dfrac{1}{2}kx^2$

$W'= 4\times W$

from equation (1)

$W'= 4\times 5$

W' = 20 J

Work take for the higher setting is equal to 20 J

Hence, the correct answer is option A

4 0

3 years ago

The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward

ch4aika [34]

The object's initial velocity is equal to $-10.59\frac{m}{s}$

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

$y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}$

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

So, isolating the initial velocity, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}$

Finally, substituting and calculating, we have:

$v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}$

Hence, we have that the initial velocity of the object is $-10.59\frac{m}{s}$

Have a nice day!

7 0

4 years ago