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3 years ago

14

A 100 N child slides down a frictionless inclined plane as shown in the figure below. Which one of the following statements is t

rue concerning the normal force that the plane exerts on the child?

1 answer:

VladimirAG [237]3 years ago

4 0

Answer:

Normal force may be less than weight, if there is some force with a component that acts opposite to the weight force. ... If the pulling force is horizontal, on a horizontal surface, then the normal force is equal to the weight. This is something your chould be able to see from a free-body diagram

Explanation:

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Directions: Name 2 kitchen tools, equipment or materials you are going to

sleet_krkn [62]

Answer:

silverware and tupperware?

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4 years ago

Which of these devices is a type of aerobic exercise machine?

Vsevolod [243]

Answer:

Aerobic exercise is an important component of a healthy lifestyle. Find out more about aerobic equipment that can help keep you fit.

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5 0

3 years ago

1. What is the value of the acceleration that the car experiences? 2. What is the value of the change in velocity that the car e

tankabanditka [31]

Answer:

All the answers are solved and explained below.

Explanation:

Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.

Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).

Data Given:

m = 1000 kg

F = 2000N

t = 10s

Q1 Solution:

Acceleration = a = ?

F = ma

a = F/m

a = 2000/ 1000

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

Q2: Solution:

Change in velocity = Δv = ?

acceleration = change in velocity / time

a = Δv/t

Δv = axt

Δv = 2 x 10

Δv = 20 m/s

Q3: Solution:

Impulse = I = ?

Impulse = Force x time

I = 2000 x 10

I = 20000 Ns

Q4: Solution:

Change in Momentum = Δp = ?

Δp = mΔv

Δp = 1000 x 20

Δp = 20000 Kgm/s

Q5: Solution:

Final velocity of the car at the end of 10 seconds = vf = ?

Δp = m x Δv

Δp = m x (vf-vi)

Δp = 1000 x (vf - 0 )

20000 = 1000 x vf

vf = 20000/1000

vf = 20 m/s

Q6: Solution:

Change in momentum the car experiences as it continues at this velocity?

Δp = ?

Δp = mΔv

Δp = m x (0)

Δp = 0

Q7: Solution:

Impulse = Change in momentum

Impulse = Δp

Implulse = 0

Q8: Solution:

Change in momentum = Δp = mΔv

Δp = m(vf-vi)

Δp = 1000 x (0-20)

Δp = -20000 kgm/s

Q9: Solution:

Impulse = Δp

Impulse = -20000 Ns

Q10: Solution:

Impulse = ?

Impulse = F x t

F = impulse/t

F = -20000/4s

F = -5000 N

Q11: Solution:

F = ma

a = ?

a = F/m

a = -5000/1000

a = -5 $m/s^{2}$

6 0

3 years ago

The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d

r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let $F_1\ and\ F_2$ are two forces. The resultant of two forces acting on the same point is given by :

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}$

Where $\theta$ is the angle between two forces

When $\theta=0$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}$

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}$

When $\theta=90^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}$

$F_R=\sqrt{F_1^2+F_2^2}$

When $\theta=180^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}$

$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}$

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

8 0

3 years ago

A listener is sitting somewhere on the line between two loudspeakers that are 10 m apart. The speakers are each emitting a sine

prohojiy [21]

Answer:

57.17 Hz 114.34 Hz 285 Hz

Explanation:

The distance between the men and 1 speaker = 3.5 m

Distance between the men and second speaker = 10-3.5= 6.5 m

Here at this point there will be no sound so there will be destructive interference

Path difference $\Delta x=6.5-3.5=3$

We know that for destructive interference $\Delta x=(2m+1)\frac{\lambda }{2}=(2m+1)\frac{v}{2f}$

$3=(2m+1)\frac{v}{2f}$

$f=(2m+1)\frac{v}{6}$ here v is the speed of sound in air

So for m =0

$f=(2\times 0+1)\frac{343}{6}=57.17H$

for m =1

$f=(2\times 1+1)\frac{343}{6}=114.34Hz$

for m=2

$f=(2\times 2+1)\frac{343}{6}=285Hz$

6 0

3 years ago