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3 years ago

Describe the difference between balanced forces and action/reaction forces

Physics

2 answers:

Alecsey [184]3 years ago

5 0

Balanced forces act on the same object and Action-Reaction forces act on different objects.

NeX [460]3 years ago

3 0

Balanced forces act on the same object and Action-Reaction forces act on different objects.

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An electron is accelerated from rest by a potential difference of 412 V. It then enters a uniform magnetic field of magnitude 18

Salsk061 [2.6K]

Explanation:

Given that,

Potential difference, V = 412 V

Magnitude of magnetic field, B = 188 mT

(a) The potential energy of electron is balanced by its kinetic energy as :

$eV=\dfrac{1}{2}mv^2$

v is speed of the electron

$v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s$

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

$r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m$

Hence, this is the required solution.

7 0

3 years ago

Show your work please

nekit [7.7K]

Answer:

$V = U + at$

Explanation:

Given the following data;

Initial velocity = 0 (since the stone is starting from rest).

Final velocity = 32 m/s

Acceleration = g = 10 m/s²

Time = 3.2 seconds

To show that the speed of the stone when it hits the ground is 32 m/s, we would use the first equation of motion;

$V = U + at$

Where;

V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.

Substituting into the formula, we have;

$32 = 0 + 10*3.2$

$32 = 0 + 32$

$32 = 32$

Proven: 32 m/s = 32 m/s

7 0

3 years ago

A ball thrown vertically upward is caught by the thrower after 4.00 sec. Find the initial velocity of the ball and the maximum h

My name is Ann [436]

Answer:

Initial velocity = 39.2m/s

Maximum height is 78.4m

Explanation:

Given

$Time, t = 4s$

Solving (a): Initial Velocity

Using first law of motion:

$v = u + at$

Where

$v = final\ velocity = 0$

$u = iniital\ velocity = ??$

 $a = acceleration = -g$  [g represents acceleration due to gravity]

$t = 4$

Substitute these value in the above formula:

$v = u + at$

$0 = u - g * 4$

$0 = u - 9.8 * 4$

Take g as 9.8m/s²

$0 = u - 39.2$

$u = 39.2m/s\\$

Hence, initial velocity = 39.2m/s

Solving (b): Maximum Height

This will be solved using second equation of motion

$s = ut + \frac{1}{2}at^2$

This becomes

$s = ut - \frac{1}{2}gt^2$

Substitute values for u, t and g

$s = 39.2 * 4 - \frac{1}{2} * 9.8 * 4^2$

$s = 156.8 - 78.4$

$s = 78.4$

Hence, the maximum height is 78.4m

7 0

4 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Zigmanuir [339]

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

4 0

3 years ago

While John is traveling along a straight interstate highway, he notices that the mile marker reads 252 km. John travels until he

irinina [24]

Answer:

total distance is 147 km

and john displacement is = 89 km

Explanation:

given data

mile marker = 252 km

reach = 134 km

retrace path = 163 km

to find out

resultant displacement

solution

we know here john travel from 252 km mark

and he reach at 134 km

so he travel distance part1 = 252 - 134

distance d1 = 252 - 134 = 118 km ..............1

and

then he start travel from 134 km mark upto 163 km mark

so distance travel in part 2 is

distance d2 = 163 - 134 = 29 km ....................2

so total distance is 118 + 29 = 147 km

and john displacement is = 252 - 163 = 89 km

8 0

3 years ago