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LenKa [72]
1 year ago
12

What is the momentum of a 533 kg blimp moving east at +75 m/s

Physics
1 answer:
mylen [45]1 year ago
7 0

Answer:

39975kgm/s due east

Explanation:

Given parameters:

Mass of the blimp  = 533kg

Velocity  = +75m/s due east

Unknown:

Momentum of the body  = ?

Solution:

The momentum of a body is the amount of motion it posses.

 Momentum is the product of mass and velocity;

 

  Momentum = mass x velocity

  Insert the parameters and solve;

    Momentum  = 533 x 75  = 39975kgm/s

The momentum of the body is 39975kgm/s due east

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
3 0
2 years ago
A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel ro
Gelneren [198K]

Answer:

59cm

Explanation:

angular velocity = 0.8 rad/s

linear velocity = angular velocity * radius

                        =0.8rad/s * 5m

                        = 4 m/s

wavelength = (V + U)/F

where,

V is the velocity of the wave

U is the velocity of the source

F is the frequency of the source.

wavelength = (350 m/s + 4 m/s ) / 600 Hz

Wavelength = 0.59m or 59 cm

4 0
2 years ago
Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A
il63 [147K]

Answer:

The question is incomplete, below is the complete question

"Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?

a. A and B must be displacements.

b. A and B must have equal lengths.

c.A and B must point in opposite direction.

d. A and B must point in the same direction.

e.A and B must be at right angles to each other."

answer:

d. A and B must point in the same direction.

Explanation:

a.false:From vector analysis, all forms of vector with the same unit can be added, and we add vector component by component. hence this defile option (a).

b. false: in addition of vectors, length is not a criteria to consider before carrying out the operation, hence vectors of different lengths can be  added

c. false: this is against the claim as vectors in opposite direction can give rise to a new vector of negative value.

d. true: this is true as vector in the same direction add up to give rise to another vector.

e.false: this is not a true assumption

4 0
2 years ago
In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc
Vesna [10]

Answer:

Explanation:

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

r=distance between point of  rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gr+gr

v_0^2=5gr

v_0=\sqrt{5gr}

v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}

6 0
2 years ago
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