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Alex777 [14]
3 years ago
10

A _____ magnetic field can create an electric current.

Physics
1 answer:
bezimeni [28]3 years ago
4 0

Answer:

D.Strong

Explanation:

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The theoretical line perpindicular to the surface where a light ray hits a mirror is called the
valkas [14]
That's called the "normal" to the surface at that point.
3 0
3 years ago
Physics, calculating net force. Please work it out for me
Margarita [4]

Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg  We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N

So, the net force is 2985.125 N.

3 0
3 years ago
Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so
riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q  8q / L²

Q = 8q  (x² / L²)

so charge required = - 8q  (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0
3 years ago
The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi
SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3

\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0
3 years ago
The relationship between a cathode and an anode involves
astra-53 [7]
It involves electrons.

The cathode is the electrode where electron deficient ions move to.

While the anode is electrode where electron excess ions move to.

So the relationship between Cathode and Anode involves electrons.

C.
5 0
3 years ago
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