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timurjin [86]
3 years ago
15

A 1,155 kg truck has a velocity of 18 m/s. What is the momentum of the truck?

Physics
1 answer:
77julia77 [94]3 years ago
8 0

The momentum of truck is 20790 kg m/s

<em><u>Solution:</u></em>

Given that we have to find the momentum of truck

From given, 1155 kg truck has a velocity of 18 m/s

Therefore,

Mass = 1155 kg

Velocity = 18 m/s

<em><u>The momentum is given by formula:</u></em>

p = m \times v

Where, m is mass in kg and v is velocity in m/s

Substituting the values we get,

p = 1155 \times 18\\\\p = 20790

Thus momentum of truck is 20790 kg m/s

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During an observation, it was noticed that light diffracts as it passes through small slits in a barrier. What does this evidenc
IgorLugansk [536]

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

3 0
3 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
3 years ago
A man weighing 680N and a woman weighing 500N have the same momentum. What is the ratio of the man's kinetic energy Km to that o
svet-max [94.6K]

Answer:

\dfrac{K_m}{K_w}=0.734

Explanation:

given,

weight of the man = 680 N

weight of the woman = 500 N

mass of man = \dfrac{680}{9.8}

  M_m = 69.4 Kg

mass of woman

= \dfrac{500}{9.8}

M_w = 51 Kg

ratio of man's kinetic energy Km to that of the woman Kw

\dfrac{K_m}{K_w}=\dfrac{\dfrac{P_m^2}{2M_m}}{\dfrac{P_w^2}{2M_w}}

momentum is same

\dfrac{K_m}{K_w}=\dfrac{2M_w}{2M_m}

\dfrac{K_m}{K_w}=\dfrac{M_w}{M_m}

\dfrac{K_m}{K_w}=\dfrac{51}{69.4}

\dfrac{K_m}{K_w}=0.734

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