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nlexa [21]
2 years ago
8

Classify each of the following statements as a characteristic (a) of electric forces only, (b) of magnetic forces only, (c) of b

oth electric and magnetic forces, or (d) of neither electric nor magnetic forces. (viii) The magnitude of the force depends on the charged object's direction of motion.
Physics
1 answer:
kakasveta [241]2 years ago
5 0

(b) Magnetic forces only

When an object is placed in an electric field it experiences an electric force given by :

F_{E} = |q| × E

Similarly when it is placed in a magnetic field it experiences a magnetic force given by :

F_{B} = |q|×v×Bsinθ

When a particle is placed in electric field the particle gets accelerated.

The electric field has a direction, positive to negative. This is the direction that the electric field will cause a positive charge to accelerate. For example If a positive charge is moving in the same direction as the electric field vector the particle's velocity will increase. If it is moving in the opposite direction it will decelerate.  ( No change in direction of motion in this case)

When a particle is moved in magnetic field ,the magnetic force is perpendicular to velocity and magnetic field . Since force is perpendicular to velocity ,it only changes the direction of motion.

To know more about electric and magnetic forces kindly refer to the below link:

brainly.com/question/1594186

#SPJ4

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Why do adults make bigger splashes when they jump into swimming pools than small children?
Semenov [28]

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

5 0
3 years ago
Which component of magma includes mobile ions of silicon?
Lemur [1.5K]

Answer:

Melt.

Explanation:

When rocks melt, they do so slowly and gradually because most rocks are made of several minerals, which all have different melting points; moreover, the physical and chemical relationships controlling the melting are complex. As a rock melts, for example, its volume changes. When enough rock is melted, the small globules of melt link up and soften the rock.

Under normal conditions, mantle rock like peridotite shouldn't melt in the Earth's upper mantle. However, by adding water you can lower the melting point of the rock. Alternatively, by decompressing the rock, you can bring it to a pressure where the melting point is lower. In both cases, basalt magma will form and considering it is hotter and less dense than the surrounding rock, it will percolate towards the surface and some of that erupts.

6 0
3 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

5 0
3 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which comp
slavikrds [6]

Answer: A.

As a diver rises, the pressure on their body decreases which allows the volume of the gas to decrease.

Explanation:

The problem is that a diver, experiences an increased pressure of water compresses nitrogen and more of it dissolves into the body. Just as there is a natural nitrogen saturation point at the surface, there are saturation points under water. Those depend on the depth, the type of body tissue involved, and also how long a diver is exposed to the extra pressure. The deeper a diver go, the more nitrogen the body absorbs.

The problem is getting rid of the nitrogen once you ascend again. As the pressure diminishes, nitrogen starts dissolving out of the tissues of the diver's body, a process called "off-gassing." That results in tiny nitrogen bubbles that then get carried to the lungs and breathed out. However, if there is too much nitrogen and/or it is released too quickly, small bubbles can combine to form larger bubbles, and those can do damage to the body, anything from minor discomforts all the way to major problems and even death.

4 0
3 years ago
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