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Gwar [14]
3 years ago
13

If air resistance can be neglected, how does the acceleration of a ball that has been tossed straight upward compare with its ac

celeration if simply dropped?
Physics
1 answer:
timama [110]3 years ago
6 0
They are the same. If this is all happening on Earth, then the ball's acceleration is 9.8 m/s^2 in either case. That's the acceleration of gravity around here.
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If a hockey player starts from rest and accelerates at a rate of 2.1 m/s2 how long does it take him to skate 30m
OLga [1]
Hello!

<span>Let us apply the time function of space, in the uniform uniform motion (UUM)
</span>
Formula: S = S_{o} +  V_{o}* t + a* \frac{t^2}{2}

Data:
S (Final position) = 30 m
So (Initial Position) = 0 m
Vo (Initial velocity) = 0 m/s
t (time) = ? (in seconds)
a (acceleration) = 2.1 m/s²

Solving:
S = S_{o} + V_{o}* t + a* \frac{t^2}{2}
30 = 0 + 0*t + 2.1* \frac{t^2}{2}
30 =  \frac{2.1t^2}{2}
30*2 = 2.1t^2
60 = 2.1t^2
2.1t^2 = 60
t^2 =  \frac{60}{2.1}
t^2 \approx 28.5
t \approx  \sqrt{28.5}
\boxed{\boxed{t \approx 5.3\:s}}\end{array}}\qquad\quad\checkmark

Answer:
<span>C. 5.3s</span>
5 0
4 years ago
________ is a force acting through distance.
Andreas93 [3]
High school???
No way
It's work.
8 0
4 years ago
A car accelerates to the right from 12 m/s to 30 m/s in 6.0 s. What was its acceleration and displacement during this time inter
Scorpion4ik [409]

Answer:

Acceleration = 3m/s^2

Displacement = 126 m

Explanation:

The velocity changed from 12 to 30 m/s in 6 seconds. Using the formula for acceleration we get

(30-12)/6= 18/6= 3m/s^2

Displacement is d=.5a*t^2+v0*t

Inputting acceleration, time, and velocity leaves us with

d=1.5 * 36 + 12 * 6 = 54 + 72 = 126 m

7 0
4 years ago
Gravitational pull is determined by ____
sergij07 [2.7K]
The correct answer is mass

8 0
4 years ago
Read 2 more answers
An 3-kg object is dropped from a height of 5 m. The rock has an impact speed<br><br> of
Andrews [41]

Answer:

The rock has an impact speed  of 9.9 m/s.

Explanation:

given information:

object's mass, m = 3 kg

height, h = 5 m

in this case, the potential energy is equal to the kinetic energy

PE = KE

mgh = \frac{1}{2} mv²

where

m = mass (kg)

g = gravitational constant (9.8 m/s²)

v = velocity (m/s)

so,

mgh = \frac{1}{2} mv²

gh = \frac{1}{2} v²

v² = 2gh

v = √2gh

  = √(2)(9.8)(5)

  = 9.9 m/s

6 0
4 years ago
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