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Sholpan [36]
3 years ago
12

Which is not a form of light?

Physics
1 answer:
kifflom [539]3 years ago
8 0

Radio waves, Middle-C, and halitosis are not forms of light.

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A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?
7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

v_y=v\ sin(25)

v_y=22\ m/s\times\ sin(25)

v_y=9.29\ m/s

Hence, the vertical component of the velocity is 9.29 m/s

3 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
Physics Homework MathPhys homie if you see this pls help
cluponka [151]

Answer:

1. -8.20 m/s²

2. 73.4 m

3. 19.4 m

Explanation:

1. Apply Newton's second law to the car in the y direction.

∑F = ma

N − mg = 0

N = mg

Apply Newton's second law to the car in the x direction.

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Given μ = 0.837:

a = -(9.8 m/s²) (0.837)

a = -8.20 m/s²

2. Given:

v₀ = 34.7 m/s

v = 0 m/s

a = -8.20 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx

Δx = 73.4 m

3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.

d = v₀t

d = (34.7 m/s) (0.56 s)

d = 19.4 m

6 0
3 years ago
A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900
sweet-ann [11.9K]

Answer:

0.16 m

Explanation:

A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).

50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³

The volume of the rectangular tank is:

volume = width × length × depth

depth = volume / width × length

depth = 0.074 m³ / 0.500 m × 0.900 m

depth = 0.16 m

3 0
3 years ago
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
Darya [45]

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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