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77julia77 [94]
3 years ago
8

A car is traveling south at a speed of 54 miles per hour and then begins traveling at a speed of 55 miles per hour but continues

traveling in the same direction. Which of the following has changed?
A. mass
B. volume
C. velocity
D. density
Physics
1 answer:
Kazeer [188]3 years ago
7 0

Explanation:

VELOCITY: BECAUSE ITS A VECTOR QUANTITY

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Main method of energy transfer​
tatyana61 [14]

Answer:

Electrification induction

6 0
3 years ago
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A bird flies 2 km to the south then 0.2 km to the north . What is the displacement of the bird?
lidiya [134]
The displacement of the bird is 1.8 km...because displacement is the distanse of the object from mean position to the travelled position.....hope it helps
8 0
3 years ago
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A 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. Calculate the
kumpel [21]

The car’s momentum after 4.21s is 24617.4 kgm/s

<h3>Newton's Second Law of Motion.</h3>

Newton's second law state that, the rate of change of momentum, is directly proportional to the applied force.

Given that a 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. To calculate the car’s momentum after 4.21 s, Let us first list all the parameters involved.

  • Velocity u = 10.2 m/s
  • Acceleration a =  2.45 m/s²
  • Mass m = 1200Kg
  • Time t = 4.21 s

From Newton's second law,

F = (mv - mu) / t

ma = (mv - mu) / t

Substitute all the parameters into the formula above.

1200 × 2.45 = ( mv - 1200 × 10.2 ) / 4.21

2940 = ( mv - 12240 ) / 4.21

Cross multiply

12377.4 = mv - 12240

Make mv the subject of the formula

mv = 12377.4 + 12240

mv = 24617.4 kgm/s

Therefore, the car’s momentum after 4.21s is 24617.4 kgm/s

Learn more about Momentum here: brainly.com/question/25121535

#SPJ1

3 0
1 year ago
Why is an alarm clock going "off" when it actually turns on?
goblinko [34]
The alarm is going off because the alarm has already been set to activate.
3 0
3 years ago
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if a 5-kg bowling ball is projected upward witha velocity of 2.0 m/s, then what is the recoil velocity of the Earth(mass=6.0×10^
vitfil [10]

by the concept of momentum conservation we can say

if net force on a system of mass is ZERO then its momentum will remain conserved

Here a ball is projected upwards so if we take Ball + Earth as a system then total momentum of the system will remain conserved

Initially when ball is on the surface of earth the system has zero momentum and hence we can say after throwing the ball momentum of earth + ball must be zero

now using same equation we can say

P_{ball} + P_{earth} = 0

m_1v_1 + m_2v_2= 0

given that

m_1 = 5 kg

v_1 = 2 m/s

m_2 = 6 * 10^{24} kg

from above equation velocity of earth will be

v_2 = \frac{m_1 v_1}{m_2}

v_2 = \frac{5*2}{6 * 10^{24}}

v_2 = 1.67 * 10^{-24} m/s

so above will be the recoil speed of earth

3 0
3 years ago
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