A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the

kakasveta [241]

From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

volume of water = 74 mL

Mass of water = 74 g

specific heat of glass = 0.18 cal/g ∙ °C

specific heat of water = 1.0 cal/g ∙ C°

Mass of glass = x g

Total heat gained by the system = 2000.0cal

Temperature rise = 20.0°C

Heat gained by system = Heat gained by glass + Heat gained by water

Heat gained by glass = x × 0.18 × 20

Heat gained by water = 74 × 1.0 × 20

Hence;

2000 = (x × 0.18 × 20) + ( 74 × 1.0 × 20)

2000 - 1480 = (x × 0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.

Learn more: brainly.com/question/1446583

3 0

1 year ago

You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By

Fittoniya [83]

Answer:

The frequency is $f = 0.221 \ Hz$

Explanation:

From the question we are told that

The time taken for it to decay to half its original size is $t = 3.40 \ ms = 3.40 *10^{-3} \ s$

Let the voltage of the capacitor when it is fully charged be $V_o$

Then the voltage of the capacitor at time t is said to be $V = \frac{V_o}{2}$

Now this voltage can be mathematical represented as

$V = V_o * e ^{-\frac{t}{RC} }$

Where RC is the time constant

substituting values

$\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }$

$0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }$

$- \frac{0.5}{RC} = ln (0.5)$

$-\frac{0.5}{RC} = -0.6931$

$RC = 0.721$

Generally the cross-over frequency for a low pass filter is mathematically represented as

$f = \frac{1}{2 \pi * RC }$

substituting values

$f = \frac{1}{2* 3.142 * 0.72 }$

$f = 0.221 \ Hz$

7 0

2 years ago

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

If electromagnetic radiation a has a lower frequency than electromagnetic radiation b the wavelength of a is

Jobisdone [24]

Inversely proportional to its frequency. If electromagnetic radiation A has a lower frequency than electromagnetic B, then compared to B, the wavelength of A is...? - equal - shorter - longer - exactly half the length of

5 0

2 years ago

If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th

cluponka [151]

Answer:

The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275 J

W= - 125 J

W' = 50 J

W(net)= -125 + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

7 0

2 years ago