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rosijanka [135]
3 years ago
12

Give 1 real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is c

onstant. Describe how it is an example of impulse and how force and time are involved.
Physics
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket

How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period

When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned  he uses more force and the ball also spends less time on the racket to produce the same momentum

Explanation:

The impulse of a force, ΔP is given by the following formula;

ΔP = F × Δt

Where ΔP is constant, we have;

F ∝ 1/Δt

Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.

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The arrows always start at the magnet's north pole and point towards its south pole. When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. So the magnets will push apart
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What is the resistance of a 600 W kettle that draws a current of 5.0 A? please answer with steps
lukranit [14]
P=I^2 *R

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5 1
3 years ago
What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic th
asambeis [7]

Answer:

\lambda = 2.57 \times 10^{-11} m

Explanation:

Average velocity of oxygen molecule at given temperature is

v_{rms} = \sqrt{\frac{3RT}{M}}

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}

v_{rms} = 483.4 m/s

now for de Broglie wavelength we know that

\lambda = \frac{h}{mv}

\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}

\lambda = 2.57 \times 10^{-11} m

7 0
3 years ago
By looking at the relative positions of the elements calcium, Ca, fluorine, F, sulfur, S, and oxygen, O, in the Periodic Table,
SVEN [57.7K]

Answer:

id say the first option.

Explanation:

hope this helps you!

8 0
3 years ago
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider compl
pickupchik [31]

Answer:

a. 2.1 s

b.0.48 Hz

c. A=24cm

d. 72cm/s

Explanation:

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?

What are the  period,

period is the time taken for a wave particle to make one complete oscillation

a) 31 / 15 = 2.066 seconds

= 2.1 s

(b) frequency : this the number of oscillation made in one seconds.

it is also the inverse of the period.

= oscillations / time

= 15/31= 0.48 Hz

(c) amplitude : maximum displacement from the origin

amplitude = 1/2 of the difference of oscillation marks

= 1/2(57-10) = 47/2cm

23.5cm

A=24cm

(d) maximum speed of the glider?

V=ωA

angular frequency *Amplitude

V=a*pi*f*amplitude

2π x frequency x amplitude = maximum speed

= 2π x .48 x 24

=72.38 cm/s

72cm/s

5 0
3 years ago
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