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klasskru [66]
3 years ago
11

A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara

tus, including the pole, rotates about the pole’s symmetry axis. The rope makes a constant angle of 70.0° with respect to the vertical. What is the tangential speed of the ball?
Physics
1 answer:
otez555 [7]3 years ago
7 0

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

r=2.3*sin70=2.161m (ball rotates in a circle)

If the system is in equilibrium, the tension is:

Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}

Replacing:

\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}

Replacing:

v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s

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Physical anthropology studies _____. the weather the human body the physics of death the material evidence collected at a crime
Elenna [48]

Physical anthropology studies the human body. This discipline is aimed at analyzing the evolution and diversity of human beings.

<h3>What is physical anthropology?</h3>

Physical anthropology is a branch of anthropology that studies the evolution and diversity of human beings.

Physical anthropology is also called biological anthropology due to its close relation with biological sciences.

This branch of anthropology (physical anthropology) is aimed at assessing the evolution and physical variation and/or behaviors of human beings.

Learn more about physical anthropology here:

brainly.com/question/8311728

5 0
3 years ago
How do snails support and protect themselves without a skeleton
snow_tiger [21]
Surely it is their shell that protects them. They hide beneath it
3 0
3 years ago
Read 2 more answers
Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph .
Elis [28]
<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours

So he saves the time t_{s}:

t_{s}=t_{1}-t_{2}=2.07-1.85=0.22 \ hours

In minutes:

t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}

5 0
3 years ago
Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at the
yKpoI14uk [10]

Answer:

See answer below

Explanation:

Hi there,

To get started, recall the Center of Mass formula for two masses:

x_c_m = \frac{m_1x_1+m_2x_2}{m_1+m_2}  where m is mass and x is displacement <em>from the center of the shape.</em>

Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.

We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:x_c_m=\frac{(-L/2)m+(L/2)2m}{m+2m} =\frac{(L/2)(-m+2m)}{3m} \\ x_c_m=\frac{L(m)}{6m} =\frac{L}{6}

Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.

Study well and persevere.

thanks,

4 0
4 years ago
d) Provided a recording medium (film) with sensitivity of 100uJ/cm2 and a power detector with 10mm of diameter, background measu
lara31 [8.8K]
We are given with:
Sensitivity: 100uJ/cm2
Diameter: 10mm
Background Optical Power: 8uW
Source Optical Power: 50uW

Required: exposure time

Solution:
Exposure time =  sensitivity x area / (background power - source power)
= 100uJ/cm2 (π/4) (1 cm)² / (80 uW - 50 uW)
= 2.62 s

The exposure time is 2.62 seconds.
3 0
3 years ago
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