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NNADVOKAT [17]
2 years ago
12

An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).

Physics
1 answer:
Anvisha [2.4K]2 years ago
4 0

Answer:

denisity = 52.33 g/cm^{3}

Explanation:

Density:

d = \frac{m}{v}

We have that m = 785 and that v = 15 cm^{3}.

d = \frac{785}{15}

d = 52.33 m^{3}

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The resistance is 5 Ω and the amount of electric current is 2 A. This means that the amount of voltage is
Hatshy [7]

Answer:

I=2A

R=5

Explanation:

formula

V=IR

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Voltage=10 volt

MARK ME BRAINLIEST THANKS MY ANSWER PLEASE

3 0
3 years ago
A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106hz. calculat
lakkis [162]
When it comes to wave behavior, there are parameters called wavelength and frequency. These two are related by speed of the radiowave. Radiowaves are electromagnetic waves which travels as fast as light. The wavelength is the distance while frequency is the reciprocal of time. When you multiply them both, you get the electromagnetic wave's speed. The equation is c = wavelength*frequency, where c is the speed of light equal to 3 x 10^8 m/s. 

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6 0
3 years ago
What is the flow rate of water in a pipe flowing full with an area of 0.3 m2 and velocity of 2.5 m/s?
sashaice [31]

Answer:

0.75 m³/s

Explanation:

Applying,

Q = vA.................... Equation 1

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From the question,

Given: v = 2.5 m/s, A = 0.3 m²

Substitute these values into equation 1

Q = 2.5(0.3)

Q = 0.75 m³/s

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4 0
2 years ago
Friction _____
kap26 [50]
<h2>A is the correct answer!</h2><h3></h3><h3>I'm too lazy to explain :(</h3><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>
6 0
2 years ago
A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force
melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}

in which \small \Delta (mV) , \small \Delta t represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by \small F_1, the following manipulation confirms the answer to this question.

\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}

Here \small F is the force that was applied to the object previously.

#SPJ4

4 0
1 year ago
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