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3 years ago

SOS HELP ME

Physics

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

Angle Vy Vy^2/(2*g)

25 6.3 2.02

30 7.5 2.87

45 10.6 5.73

50 11.5 6.75

65 13.6 9.44

70 14.1 10.1

Vy = vertical velocity = V sin theta

2 g H = Vy^2 since the final velocity is zero

H = Vy^2 / (2 * g)

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A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J

Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be $K_A$ = 1.9000000000000001 J

and the final kinetic energy from point B be $K_B$ = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

$-Q \times ( V_B - V_A) = (K_B - K_A)$

$-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)$

$15 = (K_B - 1.9000000000000001 \ J)$

$K_B = 15+ 1.9000000000000001 \ J$

$\mathbf{K_B =1 6.9000000000000001 \ J}$

3 0

3 years ago

Use wave equation calculate the speed of sound in the air if frequency of 110 hz has a wave length of 3 m

kirill115 [55]

Answer = 330 m/s

The wave equation is as follows:

Wave speed = wavelength x frequency

The known values are:
Wavelength = 3m
Frequency = 110 Hz

Substitute the known values into the wave equation to find the wave speed.

Wave speed = 3 x 110

Wave speed = 330 m/s

8 0

2 years ago

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a

seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0

3 years ago

A ball is dropped from rest. how fast is the ball going after 3 seconds

Airida [17]

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

5 0

3 years ago

A basketball player drops a 0.4-kg basketball vertically so that it is traveling at 5.7 m/s when it reaches the floor. The ball

Alik [6]

Answer:

(a) p = 3.4 kg-m/s (b) 37.78 N.

Explanation:

Mass of a basketball, m = 0.4 kg

Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)

It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)

(a) Change in momentum = final momentum - initial momentum

p = m(v-u)

p = 0.4 (2.8-(-5.7))

p = 3.4 kg-m/s

(b) Impulse = change in momentum

Ft = 3.4

We have, t = 0.09 s

$F=\dfrac{3.4}{0.09}\\\\F=37.78\ N$

Hence, this is the required solution.

4 0

3 years ago