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Alecsey [184]
3 years ago
6

Which of these is the best example of translational motion?

Physics
2 answers:
Tanya [424]3 years ago
6 0
B, air blowing from across the field is as a bullet fired from a rifle
stich3 [128]3 years ago
5 0
Would definitely be letter b, it makes the most sense
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A 3.0 X 10^3 grams rock swings in a circle with a diameter of 1000 cm. Given the constant speed around the circle is 17.90 mph (
marissa [1.9K]

Answer:

a = 12.8 m/s^2

Explanation:

To find the centripetal acceleration you use the following formula:

a_c=\frac{v^2}{r}   (1)

v: tangential speed of the rock = 17.90mph

r: radius of the orbit = 1000cm/2 = 500cm = 0.5m

You first change the units of the tangential velocity in order to replace the values of v and r in the equation (1):

17.90mph*\frac{1609.34m}{1\ m}*\frac{1\ h}{3600\ s}=8m/s

Then, you use the equation (1):

a_c=\frac{(8m/s)^2}{5m}=12.8\frac{m}{s^2}

hence, the centripetal acceleration is 12.8 m/s^2

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3 years ago
If a baseball has a negative velocity and a negative acceleration, its speed is
Korvikt [17]
Answer:constant cause it keeps happening Or it might be decreasing but I’m not sure
6 0
3 years ago
2 Magnetism comes from the word
AysviL [449]

Answer:

A Magnesia

Explanation:

The word magnetism comes from the word Magnesia, which is the name for a region in Asia Minor, where fragments of Fe3O4 ore (magnetite) were found, which attracts other metal objects.

Magnetism is a physical phenomenon by which we describe the attractive or repulsive force between materials.

This phenomenon has been known for thousands of years.

4 0
3 years ago
Example of the center of the gravity<br>​
velikii [3]

Answer:

The example of the center of the gravity is the middle of a seesaw

Explanation:

I hope this will help you and plz mark me brainlist

5 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
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