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lawyer [7]
3 years ago
12

A mass (10 kg) slides

Physics
1 answer:
notsponge [240]3 years ago
5 0
Is this the diagram????

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Why is a human powered generator better than the sun (solar cells)?
zheka24 [161]

Answer:

we can not use the suns energy too effectively in power cells and with human power we can generate more energy

6 0
2 years ago
A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to us
Rina8888 [55]

Answer:

The magnetic induction of the magnetic field is  0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒  the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can  consider the current density  as the total current over the whole area:

I(enclosed) = I / (πr ²)  * πr' ²

I(enclosed) = (I* r'²)/ (r ²)  

with I =  total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire  0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0  * (I* r'²)/ (r ²)  

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant =  4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') =  μ 0  * (I* r'²)/ (r ²)  

Because the circumference of a circle is  2 π r , when we integrate over length at a distance  r ′  from the center of wire whose crossection is a circle we get  2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B =  4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T  = 0.0005293 mT

The magnetic induction of the magnetic field is  0.0005293 mT

6 0
2 years ago
What is 541.2 mg in grams
PIT_PIT [208]

Answer:

0.5142 grams.

Explanation:

I think

5 0
2 years ago
Read 2 more answers
The electric potential at the dot in the figure is 3160 V. What is charge q?
PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


5 0
2 years ago
A man pushes a heavy cart with a force exerted of 250 Newtons to keep it moving at a constant velocity. What is the kinetic fric
miv72 [106K]

It can't be less than 250 N or the cart wouldn't move at all. That means there is only 1 answer. It's between not enough info or 250 N. The answer is 250 N. If it was any more, there would be acceleration.

3 0
3 years ago
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