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Aleksandr [31]
2 years ago
7

When a spinning bike wheel is placed horizontally, hung from a pivot at one end, the axis of rotation of the wheel will swing in

a horizontal circle. In which direction does it turn?a) upwardb) downwardc) horizontally, CWd) horizontally, CCW
Physics
1 answer:
Snowcat [4.5K]2 years ago
4 0

Answer:

answer is D

Explanation:

horizontally, CCW

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A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used
AleksAgata [21]

Explanation:

Below is an attachment containing the solution.

3 0
2 years ago
A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app
Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0
2 years ago
A tennis ball of mass m=0.060 kg and speed v=25 m/s strikes a wall at a 45 angle and rebounds with the same velocity at 45°. Wha
Diano4ka-milaya [45]

To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say

p = m\Delta v

Here,

m = mass

\Delta v = Change in velocity

As we can see there are two types of velocity at the moment the object makes the impact,

the first would be the initial velocity perpendicular to the wall and the final velocity perpendicular to the wall.

That is to say,

v_i = vcos\theta

v_f = -v sin\theta

El angulo dado es de 45° y la velocidad de 25, por tanto

v_i = (25)cos(45) = 17.68m/s

v_f = -(25)sin(45) = -17.68m/s

The change of sign indicates a change in the direction of the object.

Therefore the impulse would be as

p = 0.060(-17.68-17.68)

p = -2.12kg \cdot m/s

The negative sign indicates that the pulse is in the opposite direction of the initial velocity.

3 0
2 years ago
A 4.99 m long rod of negligible weight is attached on one end to a ball joint which allows the rod to rotate in all directions.
Aleksandr [31]

Answer:

The answer is 91.18 Nm

Explanation:

Solution

Recall that

The length of the rod = 4.99 m

∅ = 26°

Force = 62.5N

Now,

T = r * F

The direction of the torque will be in horizontally northward

The torque magnitude is  T =r F sin θ

where ∅ will be the angle between r + F  θ= 163°

Therefore,

T = 4.99 * 62.5 * sin 163

T =91.18 Nm

6 0
2 years ago
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
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