Freeeeeeeeeeeeeeeeeeeeeeeee points

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m

forsale [732]

Answer:

Acceleration due to gravity will be $g=5.718m/sec^2$

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing $=\frac{145}{100}=1.45sec$

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

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

So $1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}$

$\sqrt{\frac{0.55}{g}}=0.230$

Squaring both side

${\frac{0.3025}{g}}=0.0529$

$g=5.718m/sec^2$

So acceleration due to gravity will be $g=5.718m/sec^2$

3 0

3 years ago

A grindstone in the shape of a solid disk with a diameter of 1.0 m and a mass of 50.0 kg, is rotating at 900 rev/min. A tool is

monitta

Answer:

oh for real?

Explanation:

The solubility of glucose at 30°C is

125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

8 0

2 years ago

An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t

Fofino [41]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

$P(z)=P_{0}+\rho _wgh$

Now the gauge pressure is given by

$P(z)-P_{0}=\rho _wgh$

Applying values we get

$P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa$

8 0

3 years ago

he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug

S_A_V [24]

Answer:

207.4 N

Explanation:

The torque $\tau$ on a body is

$\tau = r* F$ where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

$\tau = rF\sin \theta$

Where $\theta$ is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation $\tau = rF\sin \theta$

$\tau = LF\sin \theta$

Where L is the length of the wrench.

Making F the subject

$F = \frac{\tau }{{L\sin \theta }}$

Force required to pull the wrench is given as,

$F = \frac{\tau }{{L\sin \theta }}$

Substitute $47{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ , 25 cm for L, and 115o for $\theta$

$\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}$

6 0

2 years ago

A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl

raketka [301]

The wavelength of the note is $\lambda = 39.1 cm = 0.391 m$ . Since the speed of the wave is the speed of sound, $c=344 m/s$ , the frequency of the note is
$f= \frac{c}{\lambda}=879.8 Hz$

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where $\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m$ is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
$L= \frac{1}{2f} \sqrt{ \frac{T}{\mu} } =0.31 m$

8 0

3 years ago