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masha68 [24]
3 years ago
11

Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point

x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?
Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

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Semenov [28]

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

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Convert 2 days into second​
eduard

Answer:

<em><u>172,000 second </u></em>

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3 0
2 years ago
Read 2 more answers
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
2 years ago
I need help, please answer
Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0
3 years ago
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

T = 2 \pi \sqrt\frac{m}{k}

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0
3 years ago
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