PLEASE HELP WITH THIS QUESTION.

inessss [21]

Answer:

4 kg at 30 m/s

Explanation:

comparing in pair, the 3rd is more than the 1st, and the 4th is more than the 2nd, so it boils down to the lasts two. calculate, and im pretty sure you get 4 kg at 30 m/s...

7 0

3 years ago

Read 2 more answers

Difference between uniform circular motion and non uniform circular motion

Goryan [66]

Uniform circular motion is the motion in which an object covers equal distance in equal interval of time in a circular path and in non uniform motion object covers equal distance in unequal interval of time in a circular path

6 0

3 years ago

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

4 years ago

A leopard with a mass of 55.00 kg climbs 12.0 m up a tree. What is it’s gain in GPE

PSYCHO15rus [73]

The answer is: 6,468 j

4 0

3 years ago

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

<em>Length of the string, L = 100 cm</em>

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The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0

3 years ago

PLEASE HELP WITH THIS QUESTION. ​

PLEASE HELP WITH THIS QUESTION.