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3 years ago

Whats the answer to this question show in the picture

Physics

1 answer:

devlian [24]3 years ago

4 0

Answer:

The answer is 10 m/s

The difference between mass and weight is that mass is the amount of matter in a material while weight is a measure of how the force of gravity acts upon that mass.

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Can an ordinary object, like a motorcycle, be mass-less? Yes or No

Drupady [299]

Answer:

no.

Explanation:

because the mass of an object never changes.

4 0

3 years ago

A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =

stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

$d_{13} =\sqrt{24.68}$ * 10⁻²m = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

F₂₃x = F₂₃ = +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

$F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2} }$

$F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2} }$ *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0

3 years ago

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

6 0

3 years ago

Discuss two reasons why people find transition between school and university

dangina [55]

Answer:

Is that your answer

6 0

3 years ago

1. On a force vs. mass graph, what would be the slope of the line?

Luden [163]

1. By Newton's second law,

F = m a

so the slope of the line would represent the mass of the object.

2. If all the forces are balanced, then the object is in equilibrium with zero net force, which in turn means the object is not accelerating. So the object is either motionless or moving at a constant speed.

4 0

3 years ago