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Mice21 [21]
4 years ago
10

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

etes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?Part A -Express your answer using two significant figures.T = _________sPart B -Express your answer using two significant figures.f = _________HzPart C -Express your answer using two significant figures.A = _________cmPart D -Express your answer using two significant figures.vmax = _________cm/s
Physics
1 answer:
pickupchik [31]4 years ago
5 0

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

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A cricket ball has mass 0.155 kg. If the velocity of a bowled ball has a magnitude of 35.0 m/s and the batted ball's velocity is
FinnZ [79.3K]

Answer:

Magnitude of change in momentum = 4.65 kg.m/s

Magnitude of impulse = 4.65 kg.m/s

Magnitude of the average force applied by the bat = 1550 N

Explanation:

Mass of the cricket ball, m = 0.155 kg

Initial velocity of the ball, u = 35.0 m/s

final velocity of the ball after hitting the bat, v = 65.0 m/s

Time of contact, t = 2.00 ms = 2.00 × 10⁻³ s

Now,

Magnitude of change in momentum = Final momentum - Initial momentum

or

Magnitude of change in momentum = ( m × v ) - ( m × u )

or

Magnitude of change in momentum = ( 0.155 × 65 ) - ( 0.155 × 35 )

or

Magnitude of change in momentum = 10.075 - 5.425 = 4.65 kg.m/s

Now, Magnitude of impulse = change in momentum

thus,

Magnitude of impulse = 4.65 kg.m/s

Now,

magnitude of the average force applied by the bat = \frac{\textup{Impulse}}{\textup{Time}}

or

magnitude of the average force applied by the bat = \frac{\textup{4.65}}{\textup{3}\times\textup{10}^{-3}}

or

Magnitude of the average force applied by the bat = 1550 N

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4 years ago
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Light shines through a single slit whose width is 5.6 × 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
Mrac [35]

Answer:

462 nm

Explanation:

Given: width of the slit, d = 5.6 × 10⁻⁴ m

Distance of the screen, D = 4.0 m

Fringe width, β = 3.3 mm = 3.3 × 10⁻³ m

First dark fringe means n =1

Wavelength of the light, λ = ?

\beta = \frac{\lambda D}{d}\\ \Rightarrow \lambda = \frac{d \beta}{D} =\frac{5.6\times 10^{-4} \times 3.3 \times 10^{-3}}{4.0} = 4.62 \times 10^{-7}m = 462 nm

5 0
3 years ago
?a wire is stretched 30% what is the percentage change in resistance ​
Marat540 [252]

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

l_{o} = 1.3l, and A_{o} = \frac{A}{1.3}

So that;

R_{o} = (ρl_{o}) ÷ A_{o} = (ρ × 1.3l) ÷ (\frac{A}{1.3})

    = (1.3lρ) ÷ (\frac{A}{1.3})

    = (1.3)^{2} × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

R_{o} = 1.69R

Where R_{o} is the new resistance, l_{o} is the new length, and A_{o} is the new area after stretching the wire.

The change in resistance of the wire = R_{o} - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = \frac{0.69R}{R} × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.

3 0
3 years ago
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