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almond37 [142]
2 years ago
15

Compute the density in g/cm? of a piece of metal that has a mass of 0.450 kg and a volume of 52 cm3

Physics
1 answer:
Y_Kistochka [10]2 years ago
8 0

Answer:

Ro = 8.65 [g/cm³]

Explanation:

We must remember that density is defined as the ratio of mass to volume.

Ro=m/V

where:

m = mass = 0.450 [kg] = 450 [g]

V = volumen = 52 [cm³]

Ro = density [g/cm³]

Now replacing:

Ro = 450/52\\Ro = 8.65 [g/cm^{3} ]

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anyanavicka [17]
That is the answer to your problem

7 0
3 years ago
A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?
madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy =\frac{1}{2} mv^2, where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0
3 years ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
We know for newton's 2nd law of motion that the force on an object is equal to the mass of the object times the acceleration of
Dmitrij [34]

         Force = (mass) x (acceleration)

           5 N  =  (9 kg) x (acceleration)

Divide each side
by  9 kg :              5 N / 9 kg  =  acceleration

                           Acceleration = (5/9)  kg-meter/sec²-kg

                                             =  0.555...  m/s²   .
3 0
3 years ago
A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33
Anuta_ua [19.1K]

Answer:

\frac{v_{2}}{v_{1}}=2.

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

\bar{K}=\frac{3k_{B}T}{2}

Now, we know that \bar{K} = (1/2)m\bar{v}^{2}

Before the absorption we have:

(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2} (1)

After the absorption,

(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2} (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}

\frac{v_{2}}{v_{1}}=\sqrt{4}

Therefore the ratio will be \frac{v_{2}}{v_{1}}=2

I hope it helps you!

4 0
3 years ago
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