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3 years ago

:) What is practical machine? what is the redation between MA and VR in a practicalmachine?

1 answer:

3 0

Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both

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Please help me with my Physical Science! 50 POINTS

DaniilM [7]

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

10kg divided by 2N

7.-Walking then pushing the moving forward

-Dribbling

-Basketball is pushed but bounces back

Explanation:

6 0

4 years ago

Read 2 more answers

Two men standing on the same side of wall and at the same distance from It, such that they are 4oom apart when one fires a gun t

Mamont248 [21]

Answer:

1. 571.43m/s

2. 142.9m and 342.9m

Explanation:

1.Take the difference in time.

1.2-0.7=0.7 seconds

Take the distance between them and divide with differnce in time.

400÷0.7=571.43 seconds.

2.Take the time of the two men and divide by two.

0.5÷2= 0.25 secs

1.2÷2= 0.6 secs

multiply each with the velocity.

0.25×571.43=142.9m

0.6×571.43=342.9m

8 0

2 years ago

What measures the amount of energy in a wave ?

Sliva [168]

Frequency measure the energy in a wave.

5 0

3 years ago

The magnetic field at the center of a 1.0-cm-diameter loop is 2.5 mT.

olga nikolaevna [1]

Answer:

(a) The current in the wire is 19.89 A

(b) The distance from the wire is 0.159 cm

Explanation:

Given;

magnetic field, B = 2.5 mT

diameter of the wire, d = 1 cm

radius of the wire, r = 0.5 cm = 0.005 m

(a) The current in the wire is calculated as;

$I = \frac{2Br}{\mu_0} \\\\I = \frac{2\times 2.5 \times 10^{-3} \times 0.005 }{4\pi \times 10^{-7} } \\\\I = 19.89 \ A$

(b) The distance from the wire where the magnetic field is 2.5 mT is calculated as;

$B = \frac{\mu_0 I}{2\pi d} \\\\where;\\\\d \ is \ the \ distance \ from \ the \ wire\\\\d = \frac{\mu_0 I}{2\pi B} = \frac{4 \pi \times 10^{-7} \times 19.89}{2\pi \times 2.5 \times 10^{-3}} = 0.00159 \ m = 0.159 \ cm$