All categories

2 years ago

How much heat energy is required to raise the temperature of 1 kilogram of steel by 10°C?

Physics

1 answer:

sammy [17]2 years ago

7 0

from

∆H=MC∆T

but M=1Kg=1000g

∆T=10°C

C=0.466J/g°C

∆H=(1000×0.466×10)Joule

=4660J

explanation

When heat energy is added to a substance, the temperature will change by a certain amount. The relationship between heat energy and temperature is different for every material, and the specific heat is a value that describes how they relate.

heat energy = (mass of substance)(specific heat)(change in temperature)

Q = mc∆T

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K

You might be interested in

Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa

Alex

Answer:

$B=1.1636*10^{-3}T$

Explanation:

Given data

$d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\$

To find

Magnitude of the net magnetic field B

Solution

The magnitude of the net magnetic field can be find as:

$B=2*u\frac{I}{2\pi r}\\ B=2*(4\pi*10^{-7} )\frac{32}{2\pi (0.022/2)} \\ B=1.1636*10^{-3}T$

3 0

3 years ago

What kind of energy does a skier have standing still at the top of a hill?

Len [333]

The skier has potential because potential energy is enery that is stored or an object that is or does not move

4 0

3 years ago

A body with the inertial

Andrews [41]

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

a) average velocity

Takes the form of:

$v_a_v_g=\frac{d_t_o_t_a_l}{t}=\frac{v+v_0}{2}$ this is the literal definition of average velocity; initial plus final divided by 2.

We know total displacement and total time elapsed, so we will use the middle form of the equation:

$v_a_v_g=\frac{1640m}{40s}=41 \ m/s$

b) the final velocity

We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

$v=(2v_a_v_g)-v_o = 2(41 \ m/s) - (8 m/s) = 74 m/s\\$ this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.

c) the acceleration

There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:

$a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}$

Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!

Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

3 0

3 years ago

a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl

bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m