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ehidna [41]
3 years ago
5

When a piece of iron or mass 78 gm is put in a graduated cylinder containing 100 cm cube of water the reading of the cylinder be

come 110cm cube
Calculate the density of iron.
Physics
1 answer:
Wittaler [7]3 years ago
7 0

Answer:

Ro = 7.8 [g/cm³]

Explanation:

According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.

V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]

We now know that density is defined as the relationship between mass and volume.

Ro = m/V_{d}

where:

Ro = density [g/cm³]

m = mass = 78 [g]

Vd = displacement volume [cm³]

Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]

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A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the
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Answer:

R=2.78\ \Omega

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

P=I^2R

Put all the values to find R.

R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega

So, the resistance is equal to 2.78\ \Omega.

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What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds? Please help
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Can velocity and acceleration point in opposite direction?
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Read 2 more answers
Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
garri49 [273]

Answer:

5.2\ \text{m/s}

70^{\circ} south of east

Explanation:

v_a = 3 m/s

\theta_a = 20^{\circ} north of east

v_b = 6 m/s

\theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

x and y component of v_a

v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

6 0
3 years ago
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