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valina [46]
3 years ago
6

A continental polar air mass forms in ____

Physics
1 answer:
Sladkaya [172]3 years ago
7 0
I believe it is A bit not completely sure
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A ball moving with an initial velocity of 5 m/s comes to rest after 2s. What was the ball's acceleration?
Inga [223]

Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t        ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity   (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

a = -2.5m/s²

Therefore, the acceleration of the ball is -2.5m/s²

NB: The negative sign shows that the ball was actually decelerating.

6 0
3 years ago
Can someone help me please
enyata [817]

Answer:

I think its distance

Explanation:

when measuring how far a p.o art u can use mm

3 0
3 years ago
Read 2 more answers
A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves
MArishka [77]

Answer:

Explanation:

Given,

  • Work done by the rope 900 m/s.
  • Angle of inclination of the slope = \theta\ =\ 12^o
  • Initial speed of the skier = v = 1.0 m/s
  • Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

\therefore W_r\ =\ W_g\ =\ 900\ J

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

  • Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

4 0
2 years ago
The three forces shown act on a particle. what is the direction of the resultant of these three forces?
melisa1 [442]
Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N
in the positive direction.

y-axis) The resultant on the y-axis is
F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
\tan \alpha =  \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7
from which we find 
\alpha=35^{\circ}
7 0
3 years ago
What are some of the characteristics scientists are looking for when they search for other Earth-like planets?
Yanka [14]
The ability to sustain life
(ie water, shelter, food, basic needs)
Hope this helped!
:-)
4 0
3 years ago
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