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2 years ago

14

An electron experiences a downward force of 12.8×10-19 N while traveling in a magnetic field of 8×10-5 T west, what is the magni

tude of the velocity?

1 answer:

SashulF [63]2 years ago

4 0

Answer:

$v=10^5\ m/s$

Explanation:

Given that,

Magnetic force acting on an electron, $F=12.8\times 10^{-19}\ N$

The magnitude of the magnetic field, $B=8\times 10^{-5}\ T$

We need to find the magnitude of the velocity. We know that the magnetic force is given by :

$F=qvB$

Where

v is the velocity

So,

$v=\dfrac{F}{qB}\\\\v=\dfrac{12.8\times 10^{-19}}{1.6\times 10^{-19}\times 8\times 10^{-5}}\\\\v=10^5\ m/s$

So, the magnitude of velocity is $10^5\ m/s$ .

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regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

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3 years ago

Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a ligh

guapka [62]

Answer:

This is because the acceleration of objects due to gravity is independent of the mass of the object and is constant for all objects, therefore, all objects fall with the same speed.

Explanation:

The weight of an object or force of gravity acting on an object on the surface of earth is a product of its mass and acceleration due to gravity.

Mathematically, w = mg

where, m=mass of the object; g = acceleration due to gravity

Also, from newton's law of gravitation, gravitational force on the object ,F = GMm/r²

where G is the gravitational constant; M is mass of Earth; m is mass of object; r is the distance of separation between the object and the center of mass of the earth which is approximately the radius of earth.

Since the weight of an object is equal to the force of gravitation acting on it

W = F

mg = GMm/r²

g = GM/r²

The expression above is that of the relationship between the force of gravity acting on a body on the earth's surface, the weight of that body and the acceleration due to gravity, g.

It can be seen that the acceleration due to gravity g is independent of the mass of the object. Therefore, the acceleration of objects due to gravity is constant for all objects and all objects fall with the same speed.

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valentinak56 [21]

Answer:

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Explanation:

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