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Marat540 [252]
3 years ago
10

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from

its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position
Physics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

615 N

Explanation:

If θ is the angle from vertical and T is the rope tension

θ = arcsin (2.0 / 3.5) = 34.85°

Summing vertical forces to zero

Tcosθ - mg = 0

T = 90.0(9.81) / cos34.85 = 1,075.85 N

Summing horizontal forces to zero

F - Tsinθ = 0

F = 1075.85sin34.85 = 1075.85(2.0/3.5) = 614.772... ≈ 615 N

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Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

3 0
3 years ago
A balloon filled with helium gas has an average density of rhob = 0.22 kg/m3. The density of the air is about rhoa = 1.23 kg/m3.
MissTica

Answer:

a=g\left(\frac{\rho_a}{\rho_b}-1\right)

45.03681 m/s²

Explanation:

F_b = Buoyant force

W = Weight of the balloon

\rho_a = Density of air = 1.23 kg/m³

\rho_b = Density of balloon = 0.22 kg/m³

v_a = Volume of air

v_b = Volume of balloon

F_b=\rho_av_bg

W=\rho_bv_bg

g = Acceleration due to gravity = 9.81 m/s²

The net force acting on the balloon is

F=F_b-W\\\Rightarrow F=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=\rho_av_bg-\rho_bv_bg\\\Rightarrow \rho_bv_ba=v_bg(\rho_a-\rho_b)\\\Rightarrow a=\frac{g}{\rho_b}(\rho_a-\rho_b)\\\Rightarrow a=g\left(\frac{\rho_a}{\rho_b}-1\right)

The equation is a=g\left(\frac{\rho_a}{\rho_b}-1\right)

a=g\left(\frac{\rho_a}{\rho_b}-1\right)\\\Rightarrow a=9.81\times \left(\frac{1.23}{0.22}-1\right)\\\Rightarrow a=45.03681\ m/s^2

The acceleration of the balloon is 45.03681 m/s²

8 0
3 years ago
A 100 kg boat is floating in water. half of the boat is submerged under water. what is the weight of the boat?
Alex17521 [72]
W=MG
w is weight 
m is mass
g is gravity 
W=(100 kg)(9.8 m/s)
W= 980 N
hope this helps
7 0
3 years ago
A 75.0 kg object decelerates from a speed of 100 m/s to 50 m/s over an interval of 25 seconds. What was the net force acting upo
lukranit [14]

Assuming constant acceleration <em>a</em>, the object has undergoes an acceleration of

<em>a</em> = (50 m/s - 100 m/s) / (25 s) = -2 m/s²

Then the net force has a magnitude <em>F</em> such that, by Newton's second law,

<em>F</em> = (75.0 kg) <em>a</em>

<em>F</em> = (75.0 kg) (-2 m/s²)

<em>F</em> = -150 N

meaning the object is acted upon by a net force of 150 N in the direction opposite the initial direction in which the object is moving.

7 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
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