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3 years ago

10

1kg slab of concrete loses 12,000j of heat when it cools from 30 Celsius to 26 Celsius. Determine the specific heat capacity of

concrete.

1 answer:

OleMash [197]3 years ago

3 0

The specific heat capacity of concrete is $3.0 J/(g^{\circ}C)$

Explanation:

When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

In this problem, we have:

m = 1 kg = 1000 g is the mass of the concrete slab

$Q = -12,000 J$ is the amount of energy lost by the slab

$\Delta T = 30-26= -4^{\circ}C$ is the change in temperature of the slab

Solving the equation for $C_s$ , we find the specific heat capacity of concrete:

$C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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How would you determine the volume of an irregular shaped object.

Archy [21]

You can use the displacement method or the eureka can so basically in the displacement can what you have to do is to put some water into a measuring cylinder and measure its volume before adding the irregular shaped object and then measuring the level of water which had been displaced and then eureka can you can check online

7 0

3 years ago

Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0

1 year ago

a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d

USPshnik [31]

Answer:

The time is $t = 2.595 \ s$

The speed is $v = 25.43 \ m/s$

Explanation:

From the question we are told that

The height of the cliff is $h = 33 \ m$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} gt^2$

before the jump the persons initial velocity is u = 0 m/s

So

$33 = 0 * t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.595 \ s$

Generally from kinematic equation

$v= u + gt$

=> $v= 0 + 9.8 * 2.595$

=> $v = 25.43 \ m/s$

3 0

3 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

3 years ago

What is the power of a student that has done a work of 10 joules in 10 seconds

Alla [95]

Answer:

1 Watt

Explanation:

P=W/t

P=10/10

P=1 Watt

8 0

3 years ago