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Anettt [7]
3 years ago
11

A car with two people traveling down the road has a mass of 100 kg and a velocity of 5 m/s. The car pulls over and picks up two

people. The mass of the car is now 150 kg. What is the new momentum if the velocity stays the same?
Physics
2 answers:
Andrei [34K]3 years ago
6 0

Answer: 750 kg-m/s

Explanation:  100% correct passed it on pf test.

Oliga [24]3 years ago
4 0

Answer:

750 kg.m/s

Explanation:

Momentum is the product of mass and velocity in kg.m/s.

In this case, initial momentum⇒p=m × v where m is mass and v is velocity

initial mass of car m₁=100 kg

Initial velocity of car v₁=5 m/s

Initial momentum= 100×5= 500 kg.m/s

New mass after picking two extra people m₂= 150 kg

Velocity  stays the same v₂=v₁= 5 m/s

New momentum= m₂× v₂

p=150×5=750 kg.m/s

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An airplane propeller is rotating at 1900 rpm (rev/min).
alexandr1967 [171]

Answer:

See explanation

Explanation:

We have to convert to angular velocity in rads-1 as follows;

Angular velocity in rad/s = 2π/60 × 1900 rpm = 199 rad/s

Given that

angular velocity =angle turned /time taken

Time taken = angle turned/angular velocity

Converting 35° to radians we have;

35 × π/180 = 0.61 radians

Time taken = 0.61 radians/199 rad/s

Time taken = 0.0031 seconds

3 0
3 years ago
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.
Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

=\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}

a = 30 units    and  b = 70 units

= \sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}

=   40 units to 100 units

hence, the correct answer is option C which is 50 units.

                       

4 0
3 years ago
Betty is sitting on of her surfboard out in the ocean. She is waiting for the perfect wave to come along so she can ride it in t
givi [52]

Crest is the part of the wave which does this.

A sound wave is the sample of disturbance resulting from the movement of strength visiting through a medium, including air, water or every other liquid or stable remember as it propagates far from the supply of the sound.

The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The sound source creates vibrations in the surrounding medium. because the supply continues to vibrate the medium, the vibrations propagate far from the supply at the rate of sound, hence forming the sound wave.

A sound wave is not a transverse wave with crests and troughs, however alternatively a longitudinal wave with compressions and rarefactions.

Learn more about wave here:- brainly.com/question/1199084

#SPJ1

7 0
1 year ago
g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at t
nlexa [21]

Answer:

58.32 N

Explanation:

Area of a circle = \pir^{2}

where r is the radius of the circle.

The cylinder has a radius of 0.02 m, its area is;

A_{1} = \pir^{2}

  = \frac{22}{7} x (0.02)^{2}

  = \frac{22}{7} x 0.0004

  = 1.2571 x 10^{-3}

Area of the cylinder is 0.0013 m^{2}.

The safety valve has a radius of 0.0075 m, its area is;

A_{2} = \pir^{2}

    = \frac{22}{7} x (0.0075)^{2}

    = \frac{22}{7} x 5.625 x 10^{-5}

    = 1.7679 x 10^{-4}

Area of the valve is 0.00018 m^{2}.

From Hooke's law, the force on the safety valve can be determined by;

F = ke

F_{2}  = 950 x 0.0085

  = 8.075 N

Minimum force, F_{1}, required can be determined by;

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

\frac{F_{1} }{0.0013} = \frac{8.075}{0.00018}

F_{1} = \frac{0.0013 *8.075}{0.00018}

    = 58.32

The minimum force that must be exerted on the piston is 58.32 N.

8 0
3 years ago
A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.
Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

F_s = F_o [\frac{V}{V_S + V} ]

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz

F_o = 200 \ Hz

(b) Apply the following formula for a moving observer and a moving source;

F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz

5 0
3 years ago
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