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GaryK [48]
4 years ago
5

When you go on a hunting trip, you should leave a hunting plan with someone you trust. What information should the plan include?

Physics
2 answers:
cestrela7 [59]4 years ago
6 0

Answer:

Explanation:

The plan should include the information like where are you going for hunt, with whom you are going, when you will return. Plan should contain the specific directions and route you are going to take, so that in case of any miss-happening your family, friend or whom you trust will be able to track you.

Plan also need to contain the information of vehicle you are going to use during the trip.

bulgar [2K]4 years ago
4 0
You should put when you will leave, where you will be, and what time you will get back. 
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If the force applied to an object is not greater than the starting friction, what will happen to the object?
bonufazy [111]

Answer:

Explanation:

the object will not move as the force exerted is not sufficient enough to overcome its force of friction

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3 years ago
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Action - reaction forces are created at the ________ time and act upon __________ objects
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6 0
3 years ago
• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.
Tanya [424]

The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

<h3>Force exerted outside the wheel</h3>

The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.

∑τ = 0

  • Let the distance traveled by the load = 1.5 m
  • Let the radius of the wheel or position of the force = 0.45 m

∑τ = R(mg) - r(F)

rF = R(mg)

0.45F = 1.5(21,200 x 9.8)

F = 6.925 x 10⁵ N.

Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.

Learn more about angular momentum here: brainly.com/question/7538238

6 0
3 years ago
A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,
beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2

∴ Average intensity is 0.477 W/m²

b) Rms value

\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C

∴ Peak value of the electric field is 18.96 N/C

8 0
3 years ago
G=fr²/m,m²
Wittaler [7]
Yes, that's correct. Newton's formula for the gravitational force, stated in this form, was used by Cavendish to measure ' G '.

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6 0
4 years ago
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