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Ivan
3 years ago
15

When a force of 15 newtons is applied to a stationary chair, it starts moving. What can you say about the frictional force betwe

en the chair and the floor of the room?
Physics
1 answer:
harkovskaia [24]3 years ago
3 0
First you need to make a difference between friction while object is stationary and the friction while object is moving. Force required to start moving some object is slightly greater than force required to maintain objects movement. That means that to move a chair you need some force F1 but you can than slightly reduce force and chair will still be moving.

Now to the problem in this question: It can be said that "stationary friction force" is equal to 15 Newtons. Its also good to know that friction force between chair and floor while you are increasing your push is also increasing and is equal to force of your push. Once it reaches 15N which is it "critical value" for that chair, chair starts moving and friction force drops a little bit and now it is called friction force of moving chair.
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A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
4 years ago
Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
brilliants [131]

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0
2 years ago
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0
3 years ago
If the index of refraction for a certain glass is 1.5, and the angle of refraction is 15 degrees for a ray of light
LiRa [457]

Answer:

\theta_i = 23^o

Explanation:

The refractive index of a medium is given by the following formula:

\eta = \frac{Sin\theta_i}{Sin\theta_r}

where,

η = refractive index = 1.5

\theta_i = angle of incidence = ?

\theta_r = angle of refraction = 15°

Therefore,

1.5 = \frac{Sin\theta_i}{Sin15^o}\\\\Sin\theta_i=(1.5)Sin15^o\\\\\theta_i = Sin^{-1}(0.388)\\\\

\theta_i = 23^o

4 0
3 years ago
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 51.5 A. The resistanc
Sliva [168]

Answer:

a).Jcu= 436.44x10^{3} \frac{A}{m^{2}}

b).λcu=1.05728 \frac{kg}{m}

c).Jal= 274.66 x10^{3} \frac{A}{m^{2}}

d).λcu=0.487 \frac{kg}{m}

Explanation:

a).

ζcu=1.7x10^{-8}Ωm

ζal=2.76x10^{-8}Ωm

A=\frac{w}{R}

wcu=ζcu*l

Acu=\frac{1.7x10^{-8}*1000}{0.144} =1.18x10^{-4} m^{2}

J=\frac{I}{Acu}=\frac{51.5A}{1.18x10^{-4}m^{2}} \\J=436.44x10^{3} \frac{A}{m^{2}}

b).

mass per unit Copper

λcu=Dcu*Acu

λcu=8960 \frac{kg}{m^{3}}*1.18x10^{-4} m^{2}

λcu=1.05728 \frac{kg}{m}

c).

wal=ζal*l

Aal=\frac{2.7x10^{-8}*1000}{0.144} =0.187x10^{-3} m^{2}

J=\frac{I}{Aal}=\frac{51.5A}{0.187x10^{-3}m^{2}} \\J=274.66x10^{3} \frac{A}{m^{2}}

d).

mass per unit Aluminum

λal=Dal*Aal

λal=2600 \frac{kg}{m^{3}}*0.1875x10^{-3} m^{2}

λcu=0.487 \frac{kg}{m}

3 0
3 years ago
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