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My name is Ann [436]

3 years ago

8

a guitar string is 0.620 m long, and oscillates at 234 Hz. if a player uses his finger to shorten the string to 0.480 m, what is

the new frequency?

2 answers:

guapka [62]3 years ago

8 0

Answer:

302.25

Explanation:

Nutka1998 [239]3 years ago

4 0

Solve the x u will get 181.16

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A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it

kompoz [17]

Answer:

a) 0.303 s b) 6.77 m/s c) - 2.97 m/s d) 7.39 m/s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

b) horizontal distance traveled = horizontal velocity (Ux) × t

Ux = horizontal distance traveled / t = 2.050 / 0.303 = 6.77 m/s

c) Velocity in the vertical direction can be calculated using

Vy = Uy - gt where g is negative since initial U is zero

Vy = - 9.81 × 0.303 = - 2.97 m/s

d) the magnitude of the velocity = resultant of the the two velocity

using Pythagoras theorem

the magnitude of the velocity = √ ( 6.77² + 2.97²) = 7.39 m/s

at angle = tan^-1 ( 2.97 / 6.77) = 23.69⁰

8 0

3 years ago

How much work did the movers do (horizontally) pushing a 46.0-kgkg crate 10.5 mm across a rough floor without acceleration, if t

-BARSIC- [3]

Answer:

7.1 J

Explanation:

From the question,

Work done by the mover = work done in pushing the crate + work done against friction

W = W'+Wf................. Equation 1

W = mgd+mgμd............ Equation 2

W = mgd(1+μ)................ Equation 3

Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of friction.

Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5

constant: g = 9.8 m/s²

Substitute these values into equation 3

W = 46×9.8×0.0105(1+0.5)

W = 7.1 J

7 0

3 years ago

Suppose a square wave signal has a 65 percent duty cycle and an on-state voltage of 40 volts DC. What is the average DC voltage

Ierofanga [76]

Answer:

The voltage is $\= DC _v = 2.6 \ V$

Explanation:

From the question we are told that

The duty cycle is p = 65% = 0.65

The on - state voltage is $V = 40 \ volt$

Generally the average DC voltage is mathematically represented as

$\= DC _v = p * V$

=> $\= DC _v = 40 * 0.65$

=> $\= DC _v = 2.6 \ V$

6 0

3 years ago

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 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:

a)

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.

b)

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

c)

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