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3 years ago

If an oscillating mass has a frequency of 1.25 Hz, it makes 100 oscillations in

Physics

1 answer:

KatRina [158]3 years ago

4 0

Answer:

Time, t = 80 seconds

Explanation:

Given that,

The frequency of the oscillating mass, f = 1.25 Hz

Number of oscillations, n = 100

We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :

$f=\dfrac{n}{t}$

$t=\dfrac{n}{f}$

$t=\dfrac{100}{1.25\ Hz}$

t = 80 seconds

So, it will make 100 oscillations in 80 seconds. Hence, this is the required solution.

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A helicopter carries a 1000-kg-car suspended from a rope below it. The helicopter flies horizontally at a constant speed of 25 m

kolbaska11 [484]

Answer:

(a) Tension, T = 28.653 kN

(b) Wind resistance force, $26.925\ kN$

Solution:

As per the question:

Mass of the car, m = 1000 kg

Speed of the helicopter, v = 25 m/s

Angle made by the rope, $theta = 20^{\circ}$

Now,

(a) To calculate the tension, T in the car:

Tension along the direction of motion, $T_{h} = Tcos20^{\circ}$

Tension along the vertical direction, $T_{v} = Tsin20^{\circ}$

Now, let the force due to the wind directed in the opposite direction of the motion be $F_{W}$ and it balances the horizontal component of the tension, T.

The vertical component is balance by the weight of the car, i.e., mg that acts vertically downwards.

Now,

$T_{v} = mg$

$Tsin20^{\circ} = 1000\times 9.8$

T = 28653 N = 28.653 kN

(b) The force of the wind resistance:

$F_{W} = T_{h}$

$F_{W} = 2cos20^{\circ} = 26925\ N = 26.925\ kN$

If the angle made by the rope with the vertical is $0^{\circ}$ :

$mg = Tsin(90^{\circ} - 0^{\circ})$

$Tsin90^{\circ} = mg = 9800\ N$

The tension in the rope will be equal to the weight the car.

Wind resistance force, $F_{W} = Tcos90^{\circ} = 0\ N$

If the angle made by the rope with the vertical is $90^{\circ}$ :

$mg = Tsin(90^{\circ} - 90^{\circ})$

T = 0 N

Wind resistance force, $F_{W} = Tsin0^{\circ}$

$Tsin0^{\circ} = mg$

$F_{W} = \infty$

There will be no tension in the rope and wind resistance will be infinite.

3 0

3 years ago

What is a blue moon

Irina18 [472]

.
a phenomenon whereby the moon appears bluish owing to smoke or dust particles in the atmosphere

8 0

3 years ago

A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H

Reika [66]

Answer:

The answer to your question is:

a) t1 = 2.99 s ≈ 3 s

b) vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ? time to reach the ground

vf = ? final speed

a) h = vot + (1/2)gt²

74 = 10t + (1/2)9.81t²

4.9t² + 10t -74 = 0 solve by using quadratic formula

t = (-b ± √ (b² -4ac) / 2a

t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

t = (-10 ± √ 1550.4 ) / 9.81

t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81

t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81

t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are

no negative times.

b) Formula vf = vo + gt

vf = 10 + (9.81)(3)

vf = 10 + 29.43

vf = 39.43 m/s

4 0

3 years ago

a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s

Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

$h = v_0_yt + \frac{1}{2} gt^2$

where;

t is the time of motion
$v_0_y$ is the initial vertical velocity of the stone = 0

$h = \frac{1}{2} gt^2$

The time taken by the first stone to hit the ground is calculated as;

$t_1 = \sqrt{\frac{2h}{g} }$

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

$t_2 = \sqrt{\frac{2h}{g} } + 1$