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3 years ago

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A particle is being accelerated through space by a 10-N force. Suddenly the particle encounters a head-on second force of 10 N i

n the opposite direction. The particle with both forces acting on it_____________.

1 answer:

nikdorinn [45]3 years ago

3 0

The particle with both forces acting on it will move at constant velocity

Explanation:

We can solve this problem by applying Newton's second law of motion, which states that the net force acting on a body is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the particle in this problem, initially it has a forward force of

$F_1 = 10 N$

Later, it encounters a second additional force in the opposite direction, therefore

$F_2 = -10 N$

This means that the net force on the particle now is

$F=F_1+F_2 = +10 +(-10) = 0$

As a consequence, the acceleration of the particle is zero:

$a=0$

And this means that the particle moves with constant velocity.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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During the solstices where do earth's poles point in relation to the sun

V125BC [204]

<span>During the December solstice the declination of the Sun is at 23.5° South of the equator. During the December solstice, locations in the Northern Hemisphere experience their shortest day. The December solstice is also the first day of winter in the Northern Hemisphere. Locations in the Southern Hemisphere have their longest day on the June solstice. This date also marks the first day of summer in the Southern Hemisphere. During the June solstice the declination of the Sun is at 23.5° North of the equator. During the June solstice, locations in the Northern Hemisphere experience their longest day. The June solstice is also the first day of summer in the Northern Hemisphere. Locations in the Southern Hemisphere have their shortest day on the June solstice. This date also marks the first day of winter in the Southern Hemisphere.</span>

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4 years ago

Please answer this question.

mylen [45]

Nothings showing upppp

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3 years ago

A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

Papessa [141]

Answer:

The acceleration is $a = 3.45*10^{3} m/s^2$

Explanation:

From the question we are told that

The radius is $d = 6.5 cm = \frac{6.5}{100} = 0.065 m$

The magnitude of the magnetic field is $B = 5.5 T$

The rate at which it decreases is $\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s$

The distance from the center of field is $r = 1.5 cm = \frac{1.5}{100} = 0.015m$

According to Faraday's law

$\epsilon = - \frac{d \o}{dt}$

and $\epsilon = \int\limits {E} \, dl$

Where the magnetic flux $\o = B* A$

E is the electric field

dl is a unit length

So

$\int\limits {E} \, dl = - \frac{d}{dt} (B*A)$

${E} l = - \frac{d}{dt} (B*A)$

Now $l$ is the circumference of the circular loop formed by the magnetic field and it mathematically represented as $l = 2\pi r$

A is the area of the circular loop formed by the magnetic field and it mathematically represented as $A= \pi r^2$

So

${E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} [ - \frac{db}{dt} ]$

Substituting values

$E = \frac{0.015}{2} (24*10^{-4})$

$E = 3.6*10^{-5} V/m$

The negative signify the negative which is counterclockwise

The force acting on the proton is mathematically represented as

$F_p = ma$

Also $F_p = q E$

So

$ma = qE$

Where m is the mass of the the proton which has a value of $m = 1.67 *10^{-27} kg$

$q = 1.602 *10^{-19} C$

So

$a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}$

$a = 3.45*10^{3} m/s^2$

8 0

3 years ago

I NEED HELP ON 2 QUESTIONS PLEASEEE

tino4ka555 [31]

Answer:

2) c) give-way vessel

3) a) With one short blast

Explanation:

2) A vessel that is required to take early substantial action to ensure avoiding collision called Give way vessel

In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel

Therefore, the correct option is c) give-way vessel

3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast

Therefore, the correct option is a) With one short blast.

4 0

4 years ago

If the distance between the plates is doubled, it's capacitance...

meriva

I’m almost sure it’s b , hope this helps on cripwalker

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3 years ago