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Sergeu [11.5K]
3 years ago
10

All simple machines are variations of which two basic machines?

Physics
1 answer:
SCORPION-xisa [38]3 years ago
5 0
L<span>ever; inclined plane</span>
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A satellite is always being pulled by gravity. <br> a. True<br> b. False
Alla [95]
There's no reason for the gravitational forces between
the Earth and the satellite to suddenly disappear. 
The statement is true.
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4 years ago
A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force
SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, \theta=30^{\circ}

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

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3 years ago
How much work done when .0080 C is moved through a potential difference of 1.5 V? Use W = qV. A.
grin007 [14]

Answer:

0.012 J

Explanation:

We are given:

q = 0.0080C

Potential difference =  1.5V

W=qV

Substituting the values into the equation:

W=0.0080*1.5= 0.012J

8 0
3 years ago
Two very long uniform lines of charge are parallel and are separated by 0.300 m. each line of charge has charge per unit length
Alenkasestr [34]

linear charge density of system of two line charges is given as

\lambda = 5.20 \muC/m

now as we know that electric field due to a line charge at some distance from it is given by

E = \frac{\lambda}{2\pi \epsilon_0 r}

so here we will first find the electric field of first line charge at the position of other line charge

E = \frac{5.20 * 10^{-6}}{2 \pi * 8.85 * 10^{-12}* 0.300}

E = 312000 N/C

now as we know that

F = qE

here q = charge on the line charge system at which force is required

E = electric field on that system of charge where force is required

now we can find the charge by

q = \lambda * L

q = 5.20 * 10^{-6}* 0.05 = 0.26 * 10^{-6} C

Now using the above formula

F = qE

F = 0.26 * 10^{-6} * 312000

F = 0.0811 N

so force on the part of wire is F = 0.0811 N

8 0
3 years ago
An object that is farther from a converging lens than its focal point always has an image that is _____.
bearhunter [10]
Smaller in size (Pt. Sized)
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3 years ago
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