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Alecsey [184]
3 years ago
14

1. Na jaką wysokość wzniesie się piłka kopnięta pionowo w górę z prędkością 20 m/s ? 2. Piłka o masie 20 dag spada z wysokości 1

5 metrów. Oblicz jej prędkość w momencie uderzenia o ziemię. Ile wynosi jej energia kinetyczna na powierzchni ziemi.
Physics
2 answers:
Tresset [83]3 years ago
6 0

Answer:

its gonn abe 23

Explanation:

sergey [27]3 years ago
6 0
The most positive answer is gonna be 23.
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A _____ is a device that multiplies force or increases speed and makes work easier.
max2010maxim [7]
Machine is the answer :)
4 0
3 years ago
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At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t
STatiana [176]
I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!
4 0
3 years ago
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As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet
Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × 10^{5}  Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × 10^{5}  = 13.6 × 10³ × 9.81 × h

h =  0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × 10^{5}  Pa

6 0
3 years ago
1. What is true about all forces?a. They are unbalanced b. They involve more than one object c. They cause objects to moved. d.
Pachacha [2.7K]

the answer is most likely D

3 0
3 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
3 years ago
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