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oee [108]
2 years ago
6

A wooden block contains some nails so that its density is exactly equal to that of water. If it is placed in a tank of water and

released from rest when it is completely submerged, it will:_____.a. remain where it is released. b. rise to the surface. c. sink to the bottom.
Physics
1 answer:
belka [17]2 years ago
8 0

Answer:

The correct answer is:

a) remain where it is released

Explanation:

The concept of density seeks to measure the weight of an object in relation to its size. It is the measure of how packed together the particles of that object are. An object placed in a liquid displaces a certain volume of the liquid, based on the relative density of the object and the liquid.

If an object is less dense than a liquid in which it is placed, it displaces a smaller volume of the liquid than its volume, hence only some part of the object will be seen to be under the liquid, the other part will float.

If an object is denser than the liquid in which it is placed, it displaces a larger volume of the liquid than its own volume, making the object to sink and is submerged, sometimes to the bottom of the liquid, but mostly below the point at which it was released.

Finally, if the density of an object and the liquid into which it is submerged is the same. the object's mass per unit volume is the same as the liquid's mass per unit volume, hence the weight and force created due to density will balance and cancel each other out hence making the object to remain where it was submerged.

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A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a
Galina-37 [17]

Answer:

E   = (0.56 \times 10^8 ) r   \   \ N/c

Explanation:

Given that:

\rho_o = (10^{-3} ) \ c/m^3

R = (0.1) m

To find  the electric field for r < R by using Gauss Law

{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)

For r < R

Q_{enclosed}=(\rho) ( \pi r^2 ) l

E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}

E= \dfrac{\rho ( r)}{2 \varepsilon_o}

where;

\varepsilon_o = 8.85 \times 10^{-12}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}

E   = (0.56 \times 10^8 ) r   \   \ N/c

4 0
2 years ago
For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci
MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

z = \frac{V_o}{i_o}

now we will have

z = \frac{125}{3.20} = 39.06 ohm

Part b)

Now we also know that

\frac{R}{z} = cos\theta

\theta = 0.982 rad = 56.3 degree

now we have

\frac{R}{39.06} = cos56.3

R = 21.7 ohm

5 0
3 years ago
Which type of electromagnetic radiation carries the most energy and has the highest frequency?
jek_recluse [69]

Answer:

Gamma rays

Explanation:

Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.

8 0
3 years ago
Read 2 more answers
A rail car moving at 10m/s collides with and connect to another stationary car. What is their final velocity?
ki77a [65]

Answer:

read this it might help some

When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater momentum change.

Explanation:

6 0
3 years ago
A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a
finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$  after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y $

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

          = 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0
2 years ago
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