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insens350 [35]
2 years ago
11

Pls answer need help

Physics
1 answer:
igomit [66]2 years ago
6 0

Answer:

See below

Explanation:

Find the NET forces on the objects

A  20==>

B    0

C  30==>

D  15==>    

So biggest accel = C because it has the most force acting on it

 next is A because it has the next biggest force

    next is  D        then    B ...B has no net force acting on it  

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A truck traveling down the highway collides with a slower moving mosquito traveling in the same direction. Which of the followin
Ipatiy [6.2K]

Answer:

B. The truck and mosquito exert the same size force on each other.

Explanation:

Newton's third law (law of action-reaction) states that

"When an object A exerts a force (action) on an object B, then object B exerts an equal and opposite force (reaction) on object A"

In this case, we can call

object A = the truck

object B = the mosquito

Thereforce according to Newton's third law, the force exerted by the truck on the mosquito is equal in magnitude to the force exerted by the mosquito on the truck (and in opposite direction).

The reason for which the mosquito will experience much more damage is the fact that the mosquito's mass is much smaller than the truck's mass, and since the acceleration is inversely proportional to the mass:

a=\frac{F}{m}

the mosquito will experience a much larger deceleration than the truck, therefore much more damage.

6 0
3 years ago
Technician A says that the freezing point of a substance is the temperature at which the substance changes state from a solid to
pashok25 [27]

Answer:

Technician B is correct

Explanation:

Freezing is a method of conversion of substance in its liquid state to solid state. It is the process by which a liquid substance changes to a solid at a particular temperature.

Increasing the pressure and decreasing the temperature of a liquid increases its freezing point. For example, in other to freeze water i.e to change water to ice, it has to be kept in a fridge at a temperature lower than the temperature of the water. The essence of covering the fridge after placing the water in the fridge is to increase the pressure of the liquid hence increasing its freezing rate.

Based on the above explanation, it can be concluded that technician B is correct.

5 0
3 years ago
Which kind of air front forms when a warm air mass meets the area of a cooler air mass?
mihalych1998 [28]
<span>Answer:  "a cold front" .
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8 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
4 years ago
The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me
Ksenya-84 [330]

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × \frac{180^{o} }{\pi }

                     = 0.02 x (180/\frac{22}{7})

0.02 radians  = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = \frac{opposite}{adjacent}

So that,

Tan 1.146° = \frac{h}{10500}

⇒ h = Tan 1.146° x 10500

      = 2.21074 x 10500

      = 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

8 0
3 years ago
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