Ask Questions Action and reaction force pairs are all around you, but they aren't always obvious. Write down
a question about a situation in which you can't identify what force pairs are at work.

Answers

Answer 1

Answer:

Writing with a pencil. The pencil pushes on the paper. The paper pushes on the pencil.

Explanation:

Newton's third law.


Related Questions

A negative charge of -0.00067 C and a positive charge of 0.00096 C are separated by 0.7 m. What is the force between the two charges?

Answers

Answer:

-11,813.87N

Explanation:

According to coulombs law, the Force between the two charges is expressed as;

F = kq1q2/d²

k is the coulombs constant = 9*10⁹kg⋅m³⋅s⁻²⋅C⁻².

q1 = -0.00067 C

q2 = 0.00096 C

d  = 0.7m

Substitute into the formula:

F =  9*10^9 *  -0.00067 * 0.00096/0.7²

F = 9*10⁹*-6.7*10⁻⁴*9.6*10⁻⁴/0.49

F = -578.88*10⁹⁻⁸/0.49

F = -578.88*10/0.49

F = -5788.8/0.49

F = -11,813.87N

Hence the force between the two charges is -11,813.87N

A wagon, initially traveling at a constant 8.2 m/s, starts going down a hill that creates an acceleration of 9.0 m/s2. What is the wagon's velocity 6.6 s after it starts accelerating down the hill?

Answers

Answer:

67.6m/s

Explanation:

Given parameters:

Initial velocity  = 8.2m/s

Acceleration  = 9.0m/s²

Time  = 6.6s

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we use the expression below:

     Acceleration  = [tex]\frac{v - u }{t}[/tex]

v is the final velocity

u is the initial velocity

 t is the time taken

    9  = [tex]\frac{v - 8.2}{6.6}[/tex]  

    v - 8.2  = 59.4

    v  = 59.4 + 8.2 = 67.6m/s

A 2 kg marble moving at 4 mi./s collides into a 1 kg marble at rest. After collision, the 2 kg marble speed decreased to 2 mi./s. Calculate the velocity and speed of the 1 kg marble immediately after colliding.

Please show work and please help me as fast as possible because this is time

Please help as soon as possible please

Please I’m begging

Answers

Answer:

[tex]2\sqrt{6}[/tex]  [tex]\frac{mi}{s}[/tex]

Explanation:

Assuming there is no waste of energy:

[tex]K_{1} = K_{2}\\\frac{1}{2}m_{1}v_{1_{1}}^{2} + \frac{1}{2}m_{2}v_{2_{1}}^2 = \frac{1}{2}m_{1}v_{1_{2}}^{2} + \frac{1}{2}m_{2}v_{2_{2}}^2\\\\=> m_{1}v_{1_{1}}^{2} + m_{2}v_{2_{1}}^2 = m_{1}v_{1_{2}}^{2} + m_{2}v_{2_{2}}^2\\\\m_{1} = 2 kg, m_{2} = 1 kg, v_{1_{1}} = 4 \frac{mi}{s} , v_{2_{1}} = 0\\=> 32 = 8 + v_{2_{2}}^{2} => v_{2_{2}} = 2\sqrt{6} \frac{mi}{s}[/tex]

In the figure shown, if angle i increases slightly, angler will
angle of
reflection
angle of
incidence
a. remain the same
Oc. increase
Ob disappear
O d. decrease

Answers

Answer:

Explanation:

According to law of reflection of light ,

angle of incidence = angle of reflection

angle of incidence = i

angle of reflection = r

when angle of incidence increases to i₁ and angle of reflection becomes r₁

According to law of reflection,

i₁ = r₁

i₁ - i = r₁ - r

increase in angle of incidence = increase in angle of reflection .

So angle of reflection also increases .

Radar tells an air traffic controller that a jet is slowing as it nears the
airport. Which might represent the jet's velocity?

A. 700 h/km

B. 700 km

C. 700 km/hr

D. 700 km/h west

Answers

Choice-D is the only velocity on the list.

(It has a speed and a direction.)

The jet's velocity might be represented by  700 km/h west. Hence, option (D) is correct.

What is velocity?

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

In this question, 700 h/km, 700 km have different unit that of velocity. So, option (A) and option (B) is not correct. . 700 km/hr has same unit that of velocity.  But it does not has direction. So, option (C) is also incorrect.  700 km/h west has the same unit of velocity with define direction. So, option (D) is correct.

Hence option (D) is correct answer.

Learn more about velocity here:

https://brainly.com/question/18084516

#SPJ6

The common version of Newton’s second law of motion, EF= ma, is actually a special case that is true only when an objects _____ is constant.

-Mass
-Momentum
-Force
-Velocity

Answers

Answer:

A

Explanation:

only is true when the objects mass is constant

Answer:

A.

Explanation:

When the mass is constant. Newton's second law will work perfectly.

Fun fact you probably won't need to know this, but the original formula that Newton derived was F=change in momentum divided by change in time. Momentum will account for a change in mass. The schooling system lies to you all the time especially in physics.

Please help me
27. A 20 kg object is at rest. A 6 N force pulls to the right on
the object for 10 seconds.
A) Find the impulse on the object.
B) Find the final momentum of the object.
C) What is its final velocity?

Answers

Answer:

a. 60 N*s

b. 60 (kg*m)/s

c. 3 m/s

Explanation:

Givens:

m = 20 kg

v_i = 0 m/s

t = 10 s

F = 6 N

a) Impulse:

I = F*t

I = 6 N*10 s

I = 60 N*s

b) Momentum:

p = v*m

F = m(a)

a = F/m

a = 6 N/20 kg

a = 0.3m/s^2

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

p = 3 m/s*20 kg

p = 60 (kg*m)/s

c. Final velocity

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

The impulse, momentum and final velocity can all be obtained from Newton's second law.

Let us recall that the impulse is obtained as the product of force and time.

Impulse = Force × time

Impulse = 6 N × 10 s = 60 Ns

From Newton's second law of motion;

F.t = mv - mu

Since the object was initially at rest;

F.t = mv

Hence, the final velocity of the object is 60 Kgms-1

The final velocity is obtained from;

F.t = mv

v = F.t/m

v = 6 × 10/20

v = 3 m/s

Learn more about impulse and momentum: https://brainly.com/question/904448

its urgent plssss help me :(

Answers

Answer:

(4) The particle travels with a constant velocity until t0 and then comes to rest.

Explanation:

Uniform rectilinear motion is defined as the motion of a particle in a straight line with constant velocity, that is, there is no change in velocity over time.

And this can be determined by means of the following equation:

[tex]x=x_{o}+v*(t-t_{o})[/tex]

where:

x = final point [m]

xo = initial point [m]

v = velocity (slope) [m/s]

t = final time [s]

to = initial time [s]

So the slope in the graph gives the constant velocity

[tex]v=\frac{x-x_{o}}{t-t_{o}}[/tex]

Therefore, after the time to there is no displacement, that is the particle comes to rest.

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