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mafiozo [28]
3 years ago
7

A baseball with a mass of 0.80 kg is given an acceleration of 20.00 m/s. How much net force was applied to the ball

Physics
1 answer:
dybincka [34]3 years ago
4 0

a x m = f

.80 x 20 = 16

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Which of the following statements best describes the relationship between force and work?
viktelen [127]

a).,  b).,  and  c).  are completely false. 
There isn't a grain of truth among them.

In Physics, the technical definition of 'Work' is (force) times (distance).

7 0
3 years ago
A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What
enot [183]

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

4 0
3 years ago
To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
3 years ago
The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

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Gravity on Mars = g₂ = \frac{1}{3} g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = \frac{1}{3}  m * g₁ = \frac{1}{3} w₁

⇒Weight of probe on Mars =\frac{1}{3} Weight of probe on earth

6 0
3 years ago
Calcular la rapidez promedio
Vlada [557]

Answer:

  v_average = 15 m / s

Explanation:

The average speed can be found in two ways,

* taking the distance traveled and divide it by the time spent

* taking the velocities in each time interval and then finding the weighted average by the time fraction

              v_average = 1 / t_total ∑  v_{i}  t_{i}vi ti

Let's apply this last equation

               

Total time is

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               t = 10 + 10 = 20   min

              v_average = 10/20 10 + 10/20 20

              v_average = 10/2 + 20/2

              v_average = 15 m / s

7 0
3 years ago
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