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BabaBlast [244]
3 years ago
13

A 0.023 kg beetle is sitting on a record player 0.15 m from the center of the record. If it takes 0.070 N of force to keep the b

eetle moving in a circle on the record, what is the tangential speed of the beetle?
Physics
2 answers:
gulaghasi [49]3 years ago
6 0
Tangential speed= 0.68 m/s
dusya [7]3 years ago
6 0

Answer:

0.68 m/s

Explanation:

The centripetal force that keeps the beetle moving in circle is given by:

F=m\frac{v^2}{r}

where

m is the mass of the beetle

v is the tangential speed of the beetle

r is the distance of the beetle from the center of the record

In this problem, we know the force (F=0.070 N), the mass of the beetle (m=0.023 kg) and the distance from the center (r=0.15 m), therefore we can re-arrange the equation to find the tangential speed:

v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(0.070 N)(0.15 m)}{0.023 kg}}=0.68 m/s

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Georgia [21]

Answer:

The answer is below

Explanation:

We are to check if the statement is true of false. If it is false, we correct the statement.

Solution:

Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.

The x coordinate represent the time and the y coordinate velocity.

5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:

Acceleration = \frac{24-0}{4-0}=6\ m/s^2

This is correct.

6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:

Acceleration = \frac{12-12}{4-0}=0\ m/s^2

This is false.

Therefore the acceleration of graph B is 0 m/s².

8 0
3 years ago
What are three observations an astronaut might make while viewing Russia at night
Lunna [17]

1. Physical size of Russia compared to other countries, despite a lack of visible borders from space.

2. Part of Russia's outline would likely be obscured by the various clouds and objects in the stratosphere; this would allow the astronaut to view potential cloud and weather patterns on earth.

3. An astronaut could see outlines of Russia's geography such as mountain ranges.

Hope that it helps :)

7 0
3 years ago
Forces that act in equal and opposite directions on an object
Akimi4 [234]
These are known as balanced forces because they will not change the motion of the object, and it will remain at rest unless forces become unbalanced- meaning they would be unequal and not opposing. 
5 0
3 years ago
If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There
GarryVolchara [31]

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is ^{0}_{0}\gamma. Hence, charge on a gamma particle is also 0.

For example, ^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy

So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.

Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if  a nucleus decays by gamma decay to a daughter nucleus.

5 0
3 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
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