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Trava [24]
2 years ago
15

Help please I’ll mark as brainliest

Physics
1 answer:
kirza4 [7]2 years ago
6 0

Answer:

yes

Explanation:

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Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c
masya89 [10]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us  find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

vₓ = xg/2y

Where we assume that x and y are known.

7 0
3 years ago
0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc
weqwewe [10]

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

4 0
3 years ago
Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.
Nikolay [14]
I think answer should be d. Please give me brainlest let me know if it’s correct
5 0
2 years ago
Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????
rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

\frac{d\textup{r}}{\textup{dt}}  = 70 cm/min

Volume of the sphere, V = \frac{\textup{4}}{\textup{3}}\pi r^3

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, \frac{d\textup{V}}{\textup{dt}}

 = \frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}

Substituting the value of \frac{dr}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

4 0
3 years ago
TRANSFORMA: 765 mm Hg a atm
-Dominant- [34]

Presión

765

=

ATMÓSFERA

1,00658

Answer:

Explanation:

5 0
3 years ago
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