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3 years ago

Use newton's law to explain the vertical acceleration of a projectile

Physics

1 answer:

NISA [10]3 years ago

5 0

Answer:

Explained below

Explanation:

Newton's first law of motion: This law states that an object will remain at rest or continue in constant motion except it's acted upon by an external force. In projectile motion, the horizontal component of velocity will remain unchanged because we ignore air resistance since no force is acting in that horizontal direction.

Newton's second law of motion: This law states that force is the product of mass and acceleration. In projectile the force acts downwards, thus f = mg.

But g = a since internal forces will cancel out.

Thus, F = ma

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________ In many cartoon shows, a character runs off a cliff, realizes his predicament, and lets out a scream. He continues to s

IrinaVladis [17]

Answer:

Here the source is moving away from the observer so frequency will be smaller than the actual frequency and since the speed is increasing so the frequency is decreasing with time so correct answer is

D) lower than the original pitch and decreasing as he falls.

Explanation:

As we know by the Doppler's effect of sound we have

so we will have

$f = f_o(\frac{v}{v + v_s})$

so here when source moves away from the observer with a some speed then the frequency of the sound observed by the observer is smaller than the actual frequency

Here we know that the speed of the source is increasing with time as the source is falling under gravity

So we can say that the pitch of the sound will decrease with time

4 0

3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

Speed is different from velocity because

shtirl [24]

Answer:

A. velocity has a direction .. .

with magnitude too but speed has only magnitude

6 0

2 years ago

Read 2 more answers

A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen

Zinaida [17]

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

$\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s$

$N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns$

Therefore, the number of turns of wire needed is 573.8 turns

4 0

2 years ago