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Luda [366]
3 years ago
6

Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou

chdown
Physics
2 answers:
ch4aika [34]3 years ago
6 0

Answer:

2s

Explanation:

Given parameters:

Distance  = 30m

Speed  = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

    Speed  = \frac{disance }{time}  

 Time taken  = \frac{distance }{speed }   = \frac{30}{15}  = 2s

The time it takes is 2s

Anton [14]3 years ago
5 0

<u>Given:</u>

  • <u>30 meters</u>
  • <u>Speed of 15 m/s</u>

<u>We will solve:</u>

<u>Speed  = </u>\frac{distance}{time}<u></u>

<u></u>

<u> Time taken  = </u>\frac{distance}{time} <u>= </u>\frac{30m}{15m/s} <u>= 2s</u>

<u></u>

<u>The time it takes is 2s</u>

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It changes slightly during a year. The average is about 93 million miles.
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3 years ago
Two capillary tubes, made of the same substance, are lowered into a water bath. The radius of the larger tube is twice that of t
Alinara [238K]

Answer: 4.4 cm.

Explanation:

Rise of water in the smaller tube= h1=8.8 centimeter(cm)

Radius of the smaller tube= r

Rise of water in the larger tube= h2 (in centimeter).

The radius of the larger tube is twice that of the smaller tube means that;

The radius in the larger tube is 2r ( 2 multiply by the radius r, of the smaller tube)

Using Jurin's law;

height or rise of liquid is inversely proportional to its radius, r.

That is; hr= constant.

Therefore, we have;

h1 × r1 = h2 × r2.

Rise in smaller tube × radius of the smaller tube = height of the larger tube × radius of the larger tube.

8.8 cm × r = h2 × 2r

= (8.8cm)r = (h2) 2r

Divide both sides by 2r, we then have;

8.8cm r/ 2r = h2

h2= 4.4cm

Therefore, the height or rise in large tube is half of that of the smaller tube.

8 0
3 years ago
What’s the answer???
kifflom [539]

Answer:

c

Explanation:e

8 0
3 years ago
Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the
svetoff [14.1K]

Answer:

\mu_k=0.18

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

x: T+F-f_k=0\\\\y:N-mg=0

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since f_k=\mu_k N and N=mg, we can rewrite the first equation as:

T+F-\mu_k mg=0

Now, we solve for \mu_k and calculate it:

\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

6 0
3 years ago
Find the interest. A 105-day note for $14,200 at 8.5% interest.
mr Goodwill [35]
<h3><u>Answer</u>;</h3>

$347.22

<h3><u>Explanation</u>;</h3>

Principal = $14,200

Rate = 8.5%

Time = 105 days = 105/365

Interest = Principal x Rate x Time

Interest = 14,200 x 0.085 x 105/365

Interest = 347.219

             = $347.22

7 0
3 years ago
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