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levacccp [35]
3 years ago
13

A soccer player heads the ball and sends it flying vertically upwards at a speed of 18.0 m/s . How high above the players ' head

does the ball travel ?​
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

16.53 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 18.0 m/s.

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

The maximum height reached by the ball can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 18² – (2 × 9.8 × h)

0 = 324 – 19.6h

Rearrange

19.6h = 324

Divide both side by 19.6

h = 324 / 19.6

h = 16.53 m

Therefore, the maximum height reached by the ball is 16.53 m

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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____
valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0
4 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

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9 0
4 years ago
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The winch takes in cable at the constant rate of 130 mm/s. if the cylinder mass is 115 kg, determine the tension in cable 1. neg
nikitadnepr [17]
By applying Newton's second law of motion;

ma = mg - T

Where,
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Tension in the cable is 1128.15 N.
8 0
4 years ago
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