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grigory [225]
3 years ago
5

A 200 g ball is tied to a string. It is pulled to an angle of 8.0° and released to swing as a pendulum. A student with a stopwat

ch finds that 10 oscillations take 12 s. How long is the string?
Physics
1 answer:
Westkost [7]3 years ago
3 0
<h2>The string is 36 cm long.</h2>

Explanation:

Period of simple is given by the equation

              T=2\pi\sqrt{\frac{l}{g}}

It will not depend upon angle and mass of ball.

A student with a stopwatch finds that 10 oscillations take 12 s.

So we can find period for 1 oscillation

                   \texttt{Period, T = }\frac{12}{10}=1.2s

Substituting in equation of period

                    T=2\pi\sqrt{\frac{l}{g}}\\\\1.2=2\pi\sqrt{\frac{l}{9.81}}\\\\l=\frac{1.2^2\times 9.81}{4\pi^2}\\\\l=0.36m

The string is 0.36 m long.

The string is 36 cm long.

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Answer: Fuel is burned in engines to make the motor move.

( Chemical to Mechanical )

Explanation:

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4 0
3 years ago
Which of the following are density labels? <br> a. Kg/L <br> b. g/m <br> c. g/mL <br> d. cm/g
bearhunter [10]
Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance.<span> </span><span>It has standard units of kg/m^3 or g/mL. So, the best answer is option C.</span>

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3 years ago
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4 years ago
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A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a
scZoUnD [109]

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
T_h = \frac{T}{2}

Now, we can determine the velocity at which the can was launched at using the following equation:
v_f = v_i + at

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}

***vsinθ is the vertical component of the velocity.

Solve for 'v':
vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}

Now, recall that:
W = \Delta KE = \frac{1}{2}m(\Delta v)^2

Plug in the expression for velocity:
W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}

B.

We can use the same process as above, where T' = 2T and Th = T.

v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}

C.

The work done in part B is 4 times greater than the work done in part A.

\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}

4 0
2 years ago
Please I really need help on this please help thank you
maw [93]
What do you need help with?
4 0
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