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lubasha [3.4K]
2 years ago
5

Ure

Physics
1 answer:
Snowcat [4.5K]2 years ago
7 0

Answer:

2. Suoco

Explanation:

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Pls helppp. Is this right?
andreyandreev [35.5K]

Answer:

yes you are totally right

7 0
3 years ago
Read 2 more answers
Please, I need help with this question.
Jet001 [13]

The x and y components of the velocity vector is 17.32 m/s and 10 m/s respectively.

<h3>What is the x - component of the velocity?</h3>

The x-component of the ball's velocity is the velocity of the ball in the horizontal direction or x-axis.

The velocity of the ball in x-direction is calculated as follows;

Vx = V cosθ

where;

  • Vx is the horizontal velocity of the ball
  • V is the speed of the ball
  • θ is the angle of inclination of the speed

Vx = (20 m/s) x (cos 30)

Vx = 17.32 m/s

The velocity of the ball in y-direction is calculated as follows;

Vy = V sinθ

where;

  • Vy is the vertical velocity of the ball
  • V is the speed of the ball
  • θ is the angle of inclination of the speed

Vy = 20 m/s x sin(30)

Vy = 10 m/s

Learn more about x and y components of velocity here: brainly.com/question/18090230

#SPJ1

3 0
1 year ago
A long, nonconducting cylinder (radius = 6.0 mm) has a nonuniform volume charge density given by αr2, where α = 6.2 mC/m5 and r
erastovalidia [21]

Answer: 2.80 N/C

Explanation: In order to calculate the electric firld inside the solid cylinder

non conductor we have to use the Gaussian law,

∫E.ds=Q inside/ε0

E*2πrL=ρ Volume of the Gaussian surface/ε0

E*2πrL= a*r^2 π* r^2* L/ε0

E=a*r^3/(2*ε0)

E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C

5 0
3 years ago
A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
WILL MARK BRAINLIEST!!! Is string a ferromagnetic material?
Contact [7]

Answer:

See, the string is made of nickel and steel (iron+carbon), materials that are ferromagnetic. That is, a magnet attracts guitar strings. When this ferromagnetic metal vibrates in the magnetic field of the pickup, that disturbs the red field lines which also cross through the coil (not shown).

Explanation:

4 0
3 years ago
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