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3 years ago

What is specific heat capacity?

2 answers:

8 0

Answer: It is the amount of thermal energy (heat) needed to raise the temperature of 1kg of a substance by 1°C. *It is important to state that it is per kg because of the word "specific". cliffffy4h and 3 more users found this answer helpful.

Anestetic [448]3 years ago

3 0

Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 kilogram of a substance by 1 kelvin (SI unit of specific heat capacity J kg−1 K−1). From: Basic Physics and Measurement in Anaesthesia (Fourth Edition), 1995.

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How does weak background radiation coming from every direction in the sky support the big bang theory

postnew [5]

Answer:

Option 1

It provides evidence of universe expansion

Explanation:

The Big Bang Theory actually explains the universe as originating from a single point, which expanded as billions of years went by.

The background radiation supports this theory in the sense that if the universe is expanding, this means that the universe is cooling and continuously losing heat, which results in background radiation

3 0

3 years ago

Which table is it I’ll mark brainlest

alexira [117]

Answer:

we need the graph to answer the question.

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3 years ago

The compass of an airplane indicates that it is headed due north and its airspeed indicator shows that it is moving through the

yarga [219]

Answer:

Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east

Explanation:

This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.

Let's take the north direction with the Y axis and the east direction as the X axis

Vy = 240 km / h airplane

Vx = 100 Km / h wind

a) See the annex

Analytical calculation of the magnitude of the speed and direction of the aircraft

V² = Vx² + Vy²

V = √ (240² + 100²)

V = 260 km/h

Airplane speed relative to the ground is 260 km/h

Tan θ = Vy / Vx

tan θ = 100/240

θ = 22.6º

Direction from north to eastb

b) What direction should the pilot have so that the resulting northbound

Vo = 240 km/h airplane

Vox = Vo cos θ

Voy = Vo sin θ

Vx = 100 km / h wind

To travel north the speeds the x axis (East) must add zero

Vx -Vox = 0

Vx = Vox = Vo cos θ

100 = 240 cos θ

θ = cos⁻¹ (100/240)

θ = 65.7º

North to West Direction

The speed in that case would be

V² = Vx² + Vy²

To go north we must find Vy

Vy² = V² - Vx²

Vy = √( 240² - 100²)

Vy = 218.2 km / h

8 0

4 years ago

A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea

soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0

3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :