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Hatshy [7]
4 years ago
7

If an object has a volume of 2.5mL and a mass of 10g than what is the density of an object?

Physics
1 answer:
Hoochie [10]4 years ago
3 0
Density is equal to mass divided by volume, so the density of the object is 10/2.5= 4<span> g/cm</span>
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A wrestler weighs in for the first match on the moon. will the athlete weigh more or less on the moon than he does on Earth?
Firdavs [7]
The moon<span> is 1/4 the size of </span>Earth<span>, so the </span>moon's<span> gravity is much less than the </span>earth's gravity, 83.3% (or 5/6) less to be exact. Finally, "weight<span>" is a measure of the gravitational pull between two objects. So of course you would </span>weigh<span> much less on the </span>moon<span>.</span>
4 0
3 years ago
A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo
Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A)  Energy

B) MRUV

<u>Energy:</u>

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0  ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2  = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) --->  V2 = \sqrt{(2*g*(H1-H2)}

Replaced values:  V2 = 14,0 m/s

<u>MRUV:</u>

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical  axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0
4 years ago
Read 2 more answers
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
What do you think would be the reason behind hot air balloon to rise up
umka21 [38]
The high temperature of the heated air inside the balloon causes the air particles to spread out. This leads to a lower density compared to colder air surrounding the balloon.

Therefore the low density causes the balloon to rise up.
4 0
3 years ago
I need help with number 2 please!!!!! Hurry!!!!!
vovangra [49]
Balance them that's the reason why
5 0
3 years ago
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