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ikadub [295]
3 years ago
6

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 61.0 cm mark on the track. The glider compl

etes 11.0 oscillations in 31.0 s
Physics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

T = 2.82 seconds.

The frequency \mathbf{f = 0.36 \ Hz}

Amplitude A = 25.5 cm

The maximum speed of the glider is \mathbf{v = 56.87 \ rad/s}

Explanation:

Given that:

the time taken for 11 oscillations is 31 seconds ;

SO, the time taken for one oscillation is :

T = \frac{31}{11}

T = 2.82 seconds.

The formula for calculating frequency can be expressed as :

f = \frac{1}{T}

f = \frac{1}{2.82}

\mathbf{f = 0.36 \ Hz}

The amplitude is determined by using the formula:

A = \frac{d}{2}

The limits that the spring makes the oscillations are from 10 cm to 61 cm.

The distance of the glider is, d = (61 - 10 )cm = 51 cm

Replacing 51 for d in the above equation

A = \frac{51}{2}

A = 25.5 cm

The maximum speed of the glider is:

v = A \omega

where ;

\omega = \frac{2 \pi}{T}

\omega = \frac{2 \pi}{2.82}

\omega = 2.23 \ rad/s

v = A \omega

v = 25.5 *2.23

\mathbf{v = 56.87 \ rad/s}

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Tasya [4]

Answer:

p = mv, where p is momentum, m is mass, and v is velocity.

a. ) m = 12kg v = 14m/s

Momentum (p) = mv

= 12kg × 14m/s

= 168kg•m/s

b.) momentum (p) = 35 kg•m/s

velocity = 3m/s

p = mv

make m the subject

divide both sides by v

we get

m = p/v

Therefore m is

m = 35 kg•m/s / 3m/s

m = 11.67kg

Therefore the mass of the object is 11.67kg

Hope this helps

3 0
3 years ago
A heavy rock is shot upward from the edge of a vertical cliff. It leaves the edge of the cliff with an initial velocity of 12 m/
jenyasd209 [6]

Answer:

The height of the building is 88.63 m.

Explanation:

Given;

initial component of vertical velocity, v_i = 12 m/s sin 26° = 5.26 m/s

initial horizontal component of the velocity, u_i = 12 m/s cos 26° =10.786 m/s

horizontal distance traveled by the rock, x = 40.4 m

time of flight is calculated as;

x = u_i t

t = x / u_i

t = 40.4 / 10.786

t = 3.75 s

Determine the final vertical velocity of the ball;

v_f = v_i + gt\\\\v_f = 5.26 + (9.8 *3.75)\\\\v_f = 42.01 \ m/s

Determine the height of the rock;

v_f^2 = v_i^2 + 2gh\\\\h = \frac{v_f^2 - v_i^2}{2g}\\\\ h = \frac{(42.01)^2 - (5.26)^2}{2*9.8}\\\\h = 88.63 \ m

Therefore, the height of the building is 88.63 m.

5 0
2 years ago
A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th
N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0
3 years ago
Why does a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude
inna [77]

Answer:

According to your question although I think an object undergoing uniform circular motion is moving with a constant speed. Nevertheless, it is accelerating due to its change in direction. The direction of the acceleration is inwards,therefore a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude.

3 0
3 years ago
A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the
Nitella [24]

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

4 0
3 years ago
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