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antoniya [11.8K]

3 years ago

10

Spiderman has a mass of 80 kg. He knows his webbing will break if it is exposed to a 200 N force. What is the maximum height he

can swing off of? You can assume aerodynamic forces prevent Spiderman from travelling greater than 30 m/s.

1 answer:

likoan [24]3 years ago

8 0

Answer:

This depends on the writers

if they want they can make spiderman deny the laws of nature

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150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit

zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

$I=\frac{150}{30}=5 A$

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3 years ago

A long wire carries a current toward the north in a magnetic field that is directed vertically downward perpendicular to the sur

xenn [34]

The direction of the magnetic force on the wire is west.

The magnetic force acting on the moving protons acts northward in the horizontal plane. If the thumb is up (current flows vertically up), the wrapped finger will be counterclockwise.

Therefore, the direction of the magnetic field is counterclockwise. Here, the magnetic field is pointing upwards (vertical magnetic field) and the electrons are moving east. Applying Fleming's left-hand rule here, we can see that the direction of force is along the south direction.

As the change in magnetic flux increases upwards, Lenz's law indicates that the induced magnetic field of the induced current must resist and the inside of the loop must be directed downwards. Using the right-hand rule, we can see that a clockwise current is induced.

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3 0

2 years ago

Read 2 more answers

What is the definition of major plates?

lutik1710 [3]

Answer:

The outer shell of the earth, the lithosphere

Explanation:

The seven major plates are the African plate, Antarctic plate, Eurasian plate, Indo-Australian plate, North American plate, Pacific plate and South American plate.

7 0

3 years ago

A box is lowered using a rope. If the acceleration of the box, is 3.3 m/s2 (downward) and its mass is 29.1 kg, what is the magni

Stells [14]

Answer:

T=189.15 N

Explanation:

As we know that for downward motion

F acting = F (weight) - Tension T

m a = mg - T

⇒ T = m (g - a)

T = 29.1 kg ( 9.8 m/s² - 3.3 m/s²)

T=189.15 N

8 0

3 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago