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ale4655 [162]
3 years ago
15

How many minutes are each half in soccer ?

Physics
1 answer:
Komok [63]3 years ago
3 0
15 minutes each half
You might be interested in
A cyclist travels at 15 m/s during a sprint finish. What is this speed in km/h
g100num [7]

Answer:

54 km/hr

Explanation:

m/s to km/hr => 18/5

15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr

8 0
3 years ago
10. Solve the following numerical problems
frosja888 [35]

Answer:

\boxed {\boxed {\sf 120 \ Joules}}

Explanation:

Work is equal to the product of force and distance.

W=F*d

The force is 8 Newtons and the distance is 15 meters.

F= 8 \ N \\d= 15 \ m

Substitute the values into the formula.

W= 8 \ N * 15 \ m

Multiply.

W= 120 \ N*m

  • 1 Newton meter is equal to 1 Joule
  • Our answer of 120 N*m equals 120 J

W= 120 \ J

The work done is <u>120 Joules</u>

3 0
3 years ago
Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the
masya89 [10]
The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2


8 0
4 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
How does increasing the distance between charged objects affect the electric force between them? the electric force increases be
konstantin123 [22]

the electric force decreases because the distance has an indirect relationship to the force

Explanation:

The electric force between two objects is given by

F=k \frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the two objects

As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.

3 0
3 years ago
Read 2 more answers
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