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vladimir1956 [14]
3 years ago
8

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30o. The sph

ere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the magnitude of the frictional force on the sphere?
I got 21.168N and the computer told me I was wrong because I was trying to find force of the max static instead of the actual static force. Please help!
Physics
1 answer:
kondaur [170]3 years ago
8 0
M*g*sin20-f=m*a 
<span>and the rotational frame </span>
<span>f*r=I*a/r </span>
<span>where f is the force of friction, a is the translational acceleration and I is the moment of inertia of the sphere </span>

<span>combine and solve for f </span>
<span>a=g*sin20-f/m </span>
<span>and </span>
<span>f*r^2/I=g*sin20-f/m </span>
<span>or </span>
<span>f=g*sin20/(r^2/I+1/m) </span>

<span>I=2*m*r^2/5 </span>
<span>therefore </span>
<span>f=2*m*g*sin20/7</span>
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marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

E = \frac{h c}{\lambda _{o}}+K

Where, K be the work function

\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K

K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )

K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )

K = 2.13 x 10^-19 J

K = 0.53 eV

3 0
3 years ago
Which objects will likely have the greatest gravitational force between them?
Mariulka [41]

Answer:

d. Two soccer balls that are touching each other

Explanation:

Let m_1 be the mass of a tennis ball, m_2 is the mass of a soccer ball.

As the mass of a soccer ball is more than the mass of a tennis ball, so

m_2 > m_1

Let d_1 be the distance between the centers of both the balls near each other and d_2 be the distance between the centers of both the balls touching each other.

So, d_2 > d_1

The gravitational force, F, between the two objects having masses M and m and separated by distance d is

F=\frac{GMm}{d^2}

Where G is the universal gravitational constant.

As, the gravitational force is directly proportional to the product of both the masses and inversely proportional to the square of the distance between them,  so selecting the larger mass (m_2, soccer ball) separated by a lesser distance (d_2, touching) to get more gravitational force.

Therefore, there will be a larger gravitational force between them when two soccer balls touching each other.

Hence, option (d) is correct.

3 0
3 years ago
A 150 kg object and a 450 kg object are separated by 0.430 m. (a) find the net gravitational force exerted by these objects on a
Dvinal [7]
Forces are vectors. They cancel each other when in opposite directions. Use the equation F=GMm/r^2 and find the difference between the two forces since they are directly opposing each other.
4 0
3 years ago
Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det
Leokris [45]

Answer:

0.037 N/m

Explanation:

The web acts as a spring, so it obeys Hook's law:

F=kx (1)

where

F is the force exerted on the web

k is the spring constant

x is the stretching/compression of the web

In this problem, we have:

- The mass of the fly is m=15 mg=15\cdot 10^{-6} kg

- The force exerted on the web is the weight of the fly, so:

F=mg=(15\cdot 10^{-6}kg)(9.81 m/s^2)=1.47\cdot 10^{-4}N

- The stretching of the web is

x=4.0 mm=0.004 m

So if we solve eq.(1) for k, we find the spring constant:

k=\frac{F}{x}=\frac{1.47\cdot 10^{-4} N}{0.004 m}=0.037 N/m

3 0
3 years ago
At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in sim
OleMash [197]

Answer:

1.869 hours

Explanation:

T = Time period = 11.1 h

Angular frequency is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{11.1}\\\Rightarrow \omega=0.566\ rad/h

The distance moved from highest to lowest level is given by

d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d

At the ocean surface x_m=0.25d

x=x_mcos(\omega t+\phi)

\phi = Phase constant = 0 as clock is started at x_0=x_m

0.25d=0.5dcos(0.56t)\\\Rightarrow cos(0.56t)=\frac{0.25}{0.5}\\\Rightarrow 0.56t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.56}\\\Rightarrow t=1.869\ h

The time taken for the water to fall the distance is 1.869 hours

7 0
3 years ago
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