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balandron [24]
3 years ago
11

2. If Chantal needs to travel an additional 150 meters beyond Taylor's house, but she

Physics
1 answer:
nirvana33 [79]3 years ago
4 0
Awnser: 600 meters per hour
Step by step explication:
Covert 14 minutes to a fraction of an hour by dividing it by 60, because 60 is how many minutes in an hour and we need to convert it to a decimals. Then divide the distance, 150 meters by the fraction of an hour, .25 to abstain the velocity (speed) of 600 meters per hour.
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Would happen if the rate that Dark Energy is expanding the universe would increase, but if gravity increased with it so the grea
mart [117]
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7 0
3 years ago
Which of the following is true of changes in momentum?
soldi70 [24.7K]

Answer: ⇒  Answer is 3

<h2>Explanation: momentum = mass × velocity</h2>

"A small force may produce a large change in momentum by acting on a very massive object".

THEY HAVEN'T GIVEN US THE TIME PERIOD NOR THE DISTANCE TRAVELED. THEREFORE WE CANNOT ACTUALLY DECIDE IF THE FORCE IS KEPT FOR A LONG TIME OR SHORT TIME. ANYWAYS SINCE THE MASS IS GIVEN AS MASSIVE , THE  MOMENTUM SHOULD BE DEFINITELY HIGH.

WHY I SAY OTHERS ARE WRONG:

1) For a small force to give a large change in momentum, it should act for a long time interval.

2) By applying a large force for a short time interval, the change of momentum should be large.

3) Correct answer.

4) Acting over a short distance can be the same as acting over a short period of time.Therefore the distance should be large in order for a larger momentum.

I HOPE IT HELPS!

6 0
3 years ago
HURRY PLZ
Natali [406]

The answer correct is C:The cell is most likely prokaryotic ,but it could be eukaryotic since some unicellular are eukaryotic.


5 0
3 years ago
Calculate the acceleration using the formula acceleration = (final velocity - initial velocity) / time.
PtichkaEL [24]

Answer:

10 km/hr/s

Explanation:

The acceleration of an object is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

For the car in this problem:

u = 0

v=60 km/h

t = 6 s

Substituting in the equation,

a=\frac{60 km/h-0}{6s}=10 km/h/s

6 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
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