PHYSICS COLLEGE

The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?

Answers

Answer 1

Answer:

Answer:

0.00098 N

Explanation:

The weight of an object is given by:

$W=mg$

where

m is the mass of the object

g is the gravitational acceleration on the planet

In this problem, we have:

$m=0.0001 kg$ is the mass of the honeybee

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting into the equation, we find:

$W=mg=(0.0001 kg)(9.8 m/s^2)=0.00098 N$

Answer 2

Answer: weight = mg
here m = 0.000100 g = 9.80
hence weight = 0.00980 kgm/s2

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The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

Answers

Answer:

Gravity

Explanation:

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

Answer:

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_(0)(1-e^{(-t)/(RC) } )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) }) \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )$

Taking natural log on both sides,

$e^{(-t)/(RC) } = 1- (V)/(V_(0) ) \n(-t)/(RC) = ln(1-(V)/(V_(0) ) )\nt = -RCln(1 - (V)/(V_(0) ))$

$t = -RC ln (1-(V)/(V_(0) ))$ (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

= $((8.85*10^(-12))) (20*10^(-4)) )/(1*10^(-3) )$

= 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

$t = -RC ln (1-(V)/(V_(0) ))$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

When running a 100 meter race Wyatt reaches his maximum speed when he is 40 meters from the starting line, and 7 seconds have elapsed since the start of the race. Wyatt continues at this max speed for the rest of the race and is 85 meters from the starting line 12 seconds after the start of the race. What is Wyatt's max speed

Answers

Answer:

9 m/s

Explanation:

Wyatt maintains the maximum speed for the rest of the race. This motion begins when his displacement is 40 m and the time is 7 s. At time 12 s, his displacement is 85 m. Because this motion is constant-velocity, the maximum speed is given by

$v_\text{max} = (85-40)/(12-7) = (45)/(5) = 9 \text{ m/s}$

Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by

Answers

Answer:

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Explanation:

Let $\vec A = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})$ and $\vec B = 4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$ , both measured in meters. The resultant vector $\vec R$ is calculated by sum of components. That is:

$\vec R = \vec A+\vec B$ (Eq. 1)

$\vec R = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})+4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$

$\vec R = (6\cdot \cos 30^(\circ)-4\cdot \sin 30^(\circ))\,\hat{i}+(6\cdot \sin 30^(\circ)-4\cdot \cos 30^(\circ))\,\hat{j}$

$\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

A satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation? A. The resultant force is the same direction as the satellite’s acceleration. B. The gravitational force acting on the satellite is negligible. C. There is no resultant force on the satellite relative to the Earth. D. The satellite does not exert any force on the Earth.

Answers

Answer:

A. The resultant force in the same direction as the satellite’s acceleration.

Explanation:

Launching a satellite in the space and then placing it in orbit around the Earth is a complicated process but at the very basic level it works on simple principles. Gravitational force pulls the satellite towards Earth whereas it acceleration pushes it in straight line.

The resultant force of gravity and acceleration makes the satellite remain in orbit around the Earth. It is condition of free fall where the gravity is making the satellite fall towards Earth but the acceleration doesn't allow it and keeps it in orbit.

Final answer:

In a circular orbit around the Earth, the resultant force acting on a satellite is in the same direction as its acceleration.

Explanation:

In a satellite orbiting the Earth in a circular orbit, there are several forces at play. The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in its orbit. The centripetal force acts towards the center of the circular orbit, while the satellite's acceleration is directed towards the center as well. Therefore, option A is correct: the resultant force is in the same direction as the satellite's acceleration.

The gravitational force acting on the satellite is not negligible; in fact, it is crucial in providing the necessary centripetal force. Therefore, option B is incorrect.

Option C is incorrect as well. There is a resultant force acting on the satellite relative to the Earth, which is responsible for keeping the satellite in its circular orbit.

Lastly, option D is also incorrect. According to Newton's third law of motion, the satellite exerts an equal and opposite force on the Earth, keeping the Earth and the satellite in orbit around their common center of mass.

Learn more about Satellites in Circular Orbit here:

brainly.com/question/36010275

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