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3 years ago

What is the direction of the magnetic field b⃗ a at point a?

Physics

1 answer:

pogonyaev3 years ago

5 0

Answer: Use the Right -hand Rule to find the direction of the magnetic field.

Explanation: A magnetic force only occur, in a magnetic field, when a particle is moving perpendicularly relative to the magnetic field. So, all three quantities are perpendicular among themselves.

The Right-Hand Rule apply for the positive charges moving in conventional current. If the moving charge is negative, the velocity must be reversed.

Right-Hand Rule determines the direction of any or all of the three and it is as follows: With you right hand, point your index finger in the direction of the velocity, your middle finger in the direction of the magnetic field and your thumb will be pointing in the direction of the magnetic force.

Another way of using the rule is when there is a current carrying wire:

The fingers of your right hand will curl into a half circle around the wire, they will point in the direction of the field. The thumb will point in the direction of the conventional current.

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A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 13.9 revolutions/s. Find the linear speed v

Lostsunrise [7]

Answer:

the linear speed of the car is 28.83 m/s

Explanation:

Given;

radius of the car, r = 0.33 m

angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s

The linear speed of the car is calculated as;

V = ωr

V = 87.35 rad/s x 0.33 m

V = 28.83 m/s

Therefore, the linear speed of the car is 28.83 m/s

3 0

3 years ago

Which feature of a human community is similar to a niche in a biological community?

pogonyaev

Answer: C. occupation

7 0

4 years ago

Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1

zheka24 [161]

Answer:

$v_{f}=1721.1m/s$

Explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

$dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\ (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2} }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s$

5 0

3 years ago

If the San Andreas Fault moves 200 cm (2 meters) per big earthquake, and plate movement is xx cm/yr (see question above), how ma

weqwewe [10]

Answer:

Option C

80 years

Explanation:

<em>From Scientific records, the rate of movement of the plate is 0.025 meters</em>

Since the fault moves at the rate of 2 meters per year, to obtain the number of years it will takes to produce an earth quake, we will have to divide the rate of movement of the fault by the rate of movement of the plate.

Number of years for earthquake = $\frac{2}{0.025}$ = 80 years.

Hence, it will take 80 years to produce one big earthquake.

5 0

4 years ago