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dexar [7]
3 years ago
11

At a certain location the horizontal component of the earth's magnetic field is

Physics
1 answer:
Katena32 [7]3 years ago
6 0
At a certain location, the horizontal component of the earth’s magnetic field is, due north<span>. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight.

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How much work is done in lifting a 6 kg object from the ground to a height of 4m?
Gekata [30.6K]
W=mgh W=(6)(9.8)(4) W= 235.2J
8 0
3 years ago
A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo
denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

\mu _smg=610

\mu _s\times 93\times 9.8=610

\mu _s=0.669

5 0
3 years ago
The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t
spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

\lambda_0 = Threshold wavelength = 459 nm

Work function is given by

W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV

The work function W0 of this metal is 2.7067 eV

4 0
3 years ago
The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0
3 years ago
A skydiver prepares to jump out of a plane. Explain how gravity and air resistance will affect the motion of the skydiver before
ahrayia [7]
Before the skydiver opens the parachute, his velocity would be increasing greatly as much as 9.8 m/s². Opening the parachute would increase the surface area to which air may cause resistance. The skydiver then reaches his terminal velocity. 
3 0
2 years ago
Read 2 more answers
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