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lbvjy [14]
3 years ago
13

The net external force acting on an object is zero. Is it possible for the object to be traveling with a velocity that is not ze

ro if the answer is yes state whether any conditions must be placed on the magnitude and direction of velocity?
Physics
1 answer:
Rzqust [24]3 years ago
4 0

Answer:

Yes, this is according to the Newton's first law of motion.

Neither its direction nor its velocity changes during this course of motion.

Explanation:

Yes, it is very well in accordance with Newton's first law of motion for a body with no force acting on it and it travels with a non-zero velocity.

During such a condition the object will have a constant velocity in a certain direction throughout its motion. Neither its direction nor its velocity changes during this course of motion.

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. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
A laboratory experiment produces a single-slit diffraction pattern on a screen. If the slit is made narrower, the bright fringes
AlekseyPX

Answer:

fringes move farther

Explanation:

The relation between the distance between the fringes is given by

y = Dλ/d

where, D is the distance between the plane of slit and the screen

d is the width of slit, λ is the wavelength of light used.

As the slit made narrower, that means d decreases, the value of y increases, that means the fringes move farther.

8 0
2 years ago
Calculate the instantaneous speed and dis-
finlep [7]

Answer:

distance = -80m

instantaneous speed = -10m/s

Explanation:

distance = velocity × time

velocity = -10m/s²

distance = -10 × 8

= -80m

instantaneous speed = distance ÷ time

= -80 ÷ 8

= -10m/s

3 0
3 years ago
Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?
kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time  = 4s

Momentum  = 1000kgm/s

Unknown:

Force  = ?

Solution:

To solve this problem, we use Newton's second law of motion;

      Ft  = Momentum

F is the force

t is the time

So;

          F x 4 = 1000kgm/s

          F  = 250N

6 0
3 years ago
Within the range of 0°c to 4°c the specific heat of water doubles when the temperature decreases, whereas the specific heat of m
AveGali [126]
This unique behavior of water is the fact that it expands while being cooled from 4 degrees Celsius to 0 degrees Celsius. In most substances, the effect of cooling is contraction, and this is true for water as well, except in the given temperature range.
4 0
3 years ago
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