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MariettaO [177]
3 years ago
7

A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are

μs=0.5 and μk=0.4, the magnitude of the frictional force on the crate is:

Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

Fr = 20 (N)

Explanation: See atached file ( free body diagram)

As for Newton´s low

∑ Fy  =  0

-mg + N = 0        ⇒  - 40 + N = 0       ⇒ N  = 40 [Newtons]

by definition :  Fr = μs * N        ⇒  Fr = 0,5 * 40      ⇒  Fr = 20 (N)

∑ Fx  =  0   body is at rest

Fe - Fr = 0

Fr > Fe

Fr > 12 (N)  the body is at rest

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A water tank is in the shape of an inverted cone with depth 10 meters, and top radius 8 meters. Water is flowing into the tank a
Elis [28]

Answer:

The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.

The depth of the water has to be equal to 7.07 m in order to have a stationary volume.

Explanation:

In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:

0.002*h^2 = 0.1

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A magnetic field is applied to a freely floating uniform iron sphere with radius R = 2.00 mm. The sphere initially had no net ma
grigory [225]

Answer:

\omega=4.2704*10^-^5 rad/s

Explanation:

Angular Momentum Formula For atoms=L_{atom}=0.12Nm_{s}h

Where:

m_{s}h is the momentum for one atom (m_s is the spin quantum number)

N is the number of atoms=\frac{N_{A}*m}{M}

Where:

N_A is Avogadro Number

m is the mass of sphere

M is the molar mass of iron

Angular Momentum Formula For atoms will be=L_{atom}=0.12\frac{N_{A}m}{M} m_{s}h

Angular Momentum of Sphere=L_{sphere}=I\omega

where:

I=\frac{2mR^{2}}{5}

So,Angular Momentum of Sphere=L_{sphere}=\frac{2mR^{2}}{5}\omega

Angular Momentum of sphere=Angular Momentum of atoms

L_{sphere}=L_{atom}

\frac{2mR^{2}}{5}\omega=0.12\frac{N_{A}m}{M} m_{s}h

For iron, m_s =\frac{1}{2}. So above equation will become:

\omega=\frac{0.12*5*N_{A}h}{4*M*R^{2} }

Where R=2mm, M=0.0558Kg/mol (Molar Mass of iron),h=Planck's Constant/2π

\omega=\frac{0.12*5*(6.022*10^{23})(6.63*10^{-34}/2*\pi)}{4*0.0558*(2*10^{-3})^{2}}

\omega=4.2704*10^-^5 rad/s

5 0
3 years ago
A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Feliz [49]

Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m

7 0
3 years ago
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