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Montano1993 [528]
3 years ago
5

A simple compass can be made by placing a small bar magnet on a cork floating in water.

Physics
1 answer:
tankabanditka [31]3 years ago
4 0

Answer:

yes

Explanation:

You might be interested in
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0
3 years ago
What is a microwave?
bonufazy [111]

Answer:

an electromagnetic wave with a wavelength in the range 0.001–0.3 m, shorter than that of a normal radio wave but longer than those of infrared radiation. Microwaves are used in radar, in communications, and for heating in microwave ovens and in various industrial processes.

7 0
2 years ago
A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe
Sindrei [870]

Answer:

The magnitude of the magnetic field is 9.3\times 10^{-5}\ T.

Explanation:

Given that,

Charge, q=-8.5\ \mu C=-8.5\times 10^{-6}\ C

Speed of the charged particle, v=9\times 10^6\ m/s

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, F=5.9\times 10^{-3}\ N

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

F=qvB\ \sin\theta

B is the magnetic field.

B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T

So, the magnitude of the magnetic field is 9.3\times 10^{-5}\ T. Hence, this is the required solution.

4 0
3 years ago
A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0
3 years ago
Two people stand facing each other at roller skating rink then push off each other
9966 [12]

a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

a)

The momentum of an object is a vector quantity given by:

p=mv

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

p=p_1 + p_2

Which can be rewritten as

p=m_1 u_1 + m_2 u_2

where m_1,m_2 are the masses of the two people and u_1,u_2 their initial velocities.

However, the two people are initially at rest, so

u_1 = 0\\u_2 = 0

Therefore the total momentum is

p=0+0=0

b)

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

p_i = p_f

where

p_i is the total momentum of the system before

p_f is the total momentum of the system after

In this problem,

p_i = 0

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

p_f = p_i = 0

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

p_f = m_1 v_1 + m_2 v_2 (1)

where:

m_1 = 30 kg is the mass of the girl

v_1 = -5 m/s is her velocity (she moves backward, so the negative sign)

m_2 = 50 kg is the mass of the boy

v_2 is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

p_f = 0 (2)

By combining eq(1) and eq(2) we get

0=m_1 v_1 + m_2 v_2

And solving for v2 we find the velocity of the boy:

v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s

and the positive sign means he is moving forward.

d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

\Delta p = F\Delta t

where

\Delta p is the change in momentum

F is the force

\Delta t is the time during which the force is applied

In this problem:

\Delta t = 2.5 s

For the boy, the change in momentum is:

\Delta p = m_2 (v_2 - u_2)

And since

m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s

We have

\Delta p = (50)(3-0)=150 kg m/s

So, the force exerted between the boy and the girl is:

F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N

8 0
3 years ago
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