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3 years ago

How many minutes in 1 hour

Physics

2 answers:

fgiga [73]3 years ago

8 0

Answer:

Explanation:

Every 60 minutes is an hour

Alexeev081 [22]3 years ago

8 0

Answer:

60; minute in 1 HR is the answer

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Suppose a log's mass is 5kg. After burning, the mass of the ash is 1 kg. Explain what could have happened to the other 4kg.

snow_lady [41]

Mass never just disappears. The other 4kg had to go somewhere. It could have left the scene of the fire in the form of smoke particles and hot gases.

3 0

4 years ago

An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electric force actin

umka2103 [35]

Answer:

F = 91.27 N

Explanation:

Parameters given:

Charge of alpha particle, q = +2.0e = $2.0 * 1.60217662 * 10^{-19}$ = $3.204 * 10^{-19} C$

Charge of gold nucleus, Q = +79e = $79 * 1.60217662 * 10^{-19}$ = $1.266 * 10^{-17} C$

Distance between the alpha particle and gold nucleus, r = $2.0 * 10^{-14} m$

Electric force is given as:

F = $\frac{kqQ}{r^2}$

where k = Coulomb's constant

Therefore:

$F = \frac{9 * 10^9 * 3.204 * 10^{-19} * 1.266 * 10^{-17}}{(2.0 * 10^{-14})^2} \\\\\\F = 91.27 N$

The electric force acting on the alpha particle when the alpha particle is $2.0 * 10^{-14} m$ from the gold nucleus is 91.27 N.

7 0

3 years ago

Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not

d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

$I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},$

where $I_0(\text{at }1000\text{ Hz})$ is the hearing threshold at 1000 Hz.

Therefore we have the following

$0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}$

$I_1$ is the threshold at 1000Hz and $I_2$ is the threshold at 100Hz.

By exponentiating we have

$10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.$

Now dividing these two equations we get

$\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.$

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.

7 0

3 years ago

The function h = -t2 + 95 models the path of a ball thrown by a boy where h represents height, in feet, and t represents the tim

Evgesh-ka [11]

Answer:

A building where the boy is on the 12 floor

Explanation:

The equation of the height or position takes into account the initial conditions.

So the general equation is

x(t) = at+c

When you solve with the initial conditions t=0 you will find the position at the begining, so this is c the constant.

in our initial equation we have 95 as the constant. and this is a equation to find height so the initial height is 95feets.

Now a common building has a distance of 7.5 feet between every floor.

You divide 95 by 7.5 and you will find 12.6

6 0

4 years ago

A ball is thrown vertically downwards at speed vo from height h. Draw velocity vs. time & acceleration vs. time graphs. In t

yuradex [85]

Answer:

Explanation:

let the ball is thrown vertically downwards with velocity u.

So, initial velocity, = - u (downwards)

acceleration = - g (downwards)

let the velocity is v after time t.

use first equation of motion

v = u + at

- v = - u - gt

v = u + gt

So, it is a straight line having slope g and y intersept is u.

The graph I shows the velocity - time graph.

Now the value of acceleration remains constant and it is equal to - 9.8 m/s^2.

So, acceleration time graph is a starigh line parallel to time axis having slope zero.

the graph II shows the acceleration - time graph.

Use III equation of motion to find the final speed in terms height.

$v^{2}=u^{2}+2gh$

And the time is

v = u + gt

$t=\frac{v-u}{g}$

4 0

4 years ago