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1 year ago

Which of the following is an accurate statement?

1 answer:

7 0

An object's momentum will change if there is a non-zero net external force acting on it. This assertion is true.

<h3></h3><h3>What is momentum?</h3>

The momentum is defined as the product of mass and the velocity of the body. It is denoted by the letter P. It occurs due to the applied force. Its unit is Kg m/s².

p=mΔV

If there is a change in the velocity there must be a force acting on the object.

If an object is acted on by a non-zero net external force, its momentum will not remain constant. is an accurate statement.

Hence, option B is correct.

To learn more about the momentum refer to the link;

brainly.com/question/4956182

#SPJ1

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Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

2 years ago

a child is stationary on a swing. The child is given a push by a parent and the child starts swinging

Nesterboy [21]

Answer:

you havent given the full question

but im guessing momentum

momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity or the impetus gained by a moving object.

Explanation:

as the child is pushed, it gathers momentum as its weight allows it be pushed forward, and the velocity is the speed driven by the amount of force the parent pushes on the child whilst they are swinging. The momentum is the result of this action

the equation that links these factors together are

p = mv

p = momentum

m = mass

v = velocity

hope i got it right ._.

3 0

1 year ago

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

2 years ago

Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various

Alexxx [7]

Answer:

$Q_{2}=1200cm^{3}/s$

Explanation:

Given data

Q₁=200cm³/s

We know that:

$F=n\frac{vA}{l}\\$

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

n₂=6.0n₁

Q=ΔP/R

$Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s$

3 0

2 years ago

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

7 0

2 years ago