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Anestetic [448]
3 years ago
7

One person holds a spring scale while another pulls until the scale reads 32 N. What is the force that the person holding the

Physics
1 answer:
Mila [183]3 years ago
5 0

Answer:

32 N

Explanation:

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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius r = 0.330

m

Force F = 290 N

Angular acceleration \alpha  = 0.814 \frac{rad}{s^{2} }

From the formula of torque,

 Γ = I\alpha                                        (1)

 Γ = rF                                       (2)

rF = I \alpha

Find momentum of inertia I from above equation,

I = \frac{rF}{\alpha }

I = \frac{0.330 \times 290}{0.814}

I = 117.56 Kg. m^{2}

Find the momentum inertia of disk,

 I = \frac{1}{2}  Mr^{2}

M = \frac{2I}{r^{2} }

M = \frac{2 \times 117.56}{(0.330)^{2} }

M = 2159.045 Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0
3 years ago
The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o
N76 [4]

Frequency = (speed) / (wavelength)

Frequency = (331 m/s) / (0.6 m) = 551.7 Hz
3 0
3 years ago
The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro
shusha [124]

Answer:

1.6\times 10^{-18} C  

Explanation:

The fundamental charge of a single electron is 1.6\times 10^{-19} C.

If there are 10 excess electrons, the net charge that would be measured should be 10 times the fundamental charge of a single electron:

Q=nq_e\\Q= 10\times 1.6\times 10^{-19} C\\Q= 1.6 \times 10^{-18} C

3 0
3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
Which statement about cellulose is true?
prohojiy [21]
The correct answer is D
7 0
3 years ago
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