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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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Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

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3 years ago
Throughout the problem, take the speed of sound in air to be 343 .
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Answer:

A pipe has a length of 1.15 m.

a) Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz

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Answer in units of Hz

Explanation:

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3 years ago
Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the
dangina [55]

Answer:

a. The spheres will attract each other.

Explanation:

When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.

  • When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.
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How fast can Usain Bolt run if it takes him 9.9 s to run 100m?​
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Explanation:

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Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
3 years ago
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