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GalinKa [24]
2 years ago
6

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on

the surface. The sanding block weighs 1.70 N and Brenda pushes on it with a force of 3.00 N at an angle of 30.0° with respect to the vertical, and angled toward the wall. What is the coefficient of kinetic friction between the wall and the block?

Physics
1 answer:
vova2212 [387]2 years ago
3 0

Answer:

1.0193

Explanation:

The concepts that we apply here is Frictional Force with a Force on the object with a angle.

We know that Frictional Force is equal to

F_f = \mu N

The force acts downward because the moving is upward, then from the image that I attached, the forve is equal to,

Fcos30° = 3N(cos30) = 2.6

In Y we have then

1.7N + f_x = 2.6

f_x = 1.529N

For the horizontal componentes we have that

N = (3N)sin30\°

N = 1.5 N

Using the first equation we have that

F_f = \mu N

1.529N = \mu 1.5 N

Re-arrange for \mu

\mu = 1.0193

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Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
where
k= \frac{M_{a}}{M_{b}} 
7 0
2 years ago
A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil
NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

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cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ_{crss} = σ_{Y} ( cosФ cosλ)_{max}

now by putting values,

     = (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

3 0
2 years ago
Батискаф витримує тиск 60 МПа. Чи можна провести дослідження
Elanso [62]

1) Yes

2) 6.34\cdot 10^9 N

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

p=p_0+\rho g h

where

p_0 = 101,300 Pa is the atmospheric pressure

\rho is the fluid density

g=10 m/s^2 is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

\rho=1030 kg/m^3 is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

p=\frac{F}{A}

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

p=56.1\cdot 10^6 Pa is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

A=4\pi r^2

Therefore, we can find the force exerted on it by re-arranging the previous equation:

F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N

6 0
2 years ago
The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900
Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

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(b) Now horizontally,

L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.

3 0
2 years ago
Compare the momentum of a 6,300-kg elephant walking 0.11m/s and a 50 kg dolphine swimming 10.4m/s
Cerrena [4.2K]
Momentum of elephant = 6300 * 0.11 = 693 kg-m/s

Momentum for  the dolphin = 50 * 10.4 =  520 kg-m/s

Difference in momentum = 693 - 520 = 173 kg-m/s
7 0
2 years ago
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