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yulyashka [42]
2 years ago
11

Physics students study a piano being pulled across a room on a rug. They know that when it is at rest, it experiences a gravitat

ional force of 2200 N. When it is being pulled with 1200 N it is sliding across the room in equilibrium, but the students feel like they had to pull harder than 1200 N at first to begin motion. Which best describes the friction acting on the piano? The static and kinetic frictional forces are equal, so the static frictional force is equal to 1200 N. The kinetic frictional force is greater than the static frictional force, so the kinetic frictional force is greater than 1200 N. The static frictional force is greater than the kinetic frictional force, so the static frictional force is greater than 1200 N. The static and kinetic friction are both equal to the gravitational force of 2200 N.
Physics
1 answer:
evablogger [386]2 years ago
3 0

Answer:

The static frictional force is greater than the kinetic frictional force, so the static frictional force is greater than 1200 N.

Explanation:

The Frictional force is a force that acts against the motion of any object that moves on a surface, which is due to the interaction between the molecules on the object's surface and the surface molecules on which the object moves.

\mu_k < \mu_s,

As we know that by the above equation, Static friction is often stronger than kinetic friction (this is why making an object begin moving is usually harder than maintaining it in motion across a surface).

Therefore the option 3 is correct according to the given situation

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A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times
notka56 [123]

Answer:

 v_{f} = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

       v_{f} - v₀ =  v_{r} ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is    

       M = 2.5 Mf

       v_{f} - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

       v_{f} = 1,386 m / s

3 0
2 years ago
A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an
Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

\omega = 2\pi f

\omega = 2\pi (2.5)

\omega = 5\pi rad/s

The angular displacement is given as the form:

\theta (t) = \theta_0 cos(\omega t)

In the equlibrium we have to t=0, \theta(t) = \theta_0 and in the given position we have to

\theta(t) = \theta_0 cos(5\pi t)

Derived the expression we will have the equivalent to angular velocity

\frac{d\theta}{dt} = 2.7rad/s

Replacing,

\theta_0(sin(5\pi t))5\pi = 2.7

Finally

\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°

Therefore the maximum angular displacement is 9.848°

6 0
2 years ago
A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir
Semmy [17]

Answer:

The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}

\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}

\dfrac{1}{u}=\dfrac{1}{6}

u=6\ m

We need to calculate the magnification

Using formula of magnification

m=-\dfrac{v}{u}

Put the value into the formula

m=-\dfrac{0.24}{-6}

m=0.04

We need to calculate the height of the object

Using formula of magnification

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

6 0
2 years ago
Read 2 more answers
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
schepotkina [342]
If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!
5 0
2 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
2 years ago
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