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Margaret
10 days ago
15

Bjorn is holding a tennis ball outside a second floor window (3.5 meters from the ground) and billie jean is holding one outside

a third floor window (6.25 meters from the ground) how much more GPE does billie jean's tennis ball have? (each tennis ball has a mass of 0.06 kg)
Physics
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5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be
Sav [3153]

Answer:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

Tension in the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

v = \sqrt{\frac{T}{\mu} }

where;

μ represents mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}

The new tension T₂ will be computed as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Consequently, the tension in the string when the speed was increased is 134.53 N

3 0
2 months ago
A 4.0 g string, 0.36 m long, is under tension. The string produces a 500 Hz tone when it vibrates in the third harmonic. The spe
ValentinkaMS [3465]
The string vibrates in its third harmonic, where n = 3. The length of the string, l, measures 0.36 m. The frequency of the sound produced is f = 500 Hz. The speed of sound in air is 344 m/s. To find the speed of sound generated by the string in the third harmonic, we can apply the appropriate formula for frequency.
5 0
1 month ago
An object is traveling at a constant velocity of 15 m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11
serg [3582]

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

7 0
2 months ago
When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup
Sav [3153]

Answer:

W_f = 148.17J

Explanation:

The friction created between the tire and the ground generates thermal energy as force is applied during skidding.

The mentioned force relates to half the impact on the rear tire, resulting in a calculated normal force of,

N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N

The work executed is determined by the frictional force and the distance covered,

W_f = fd = \mu_k Nd

Where \mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)

The thermal energy produced from the work done is,

W_f = 148.17J

3 0
2 months ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
serg [3582]

Answer:

29.4 N/m

0.1

Explanation:

a) The restoring force is described by: F_r = -k*x.

The gravitational force is given as: F_g = mg.

Where:

F_r = restoring force,

F_g = gravitational force,

g represents acceleration due to gravity,

and k is the force constant, while x1 and x2 are the displacements for the two masses.

Given:

m1 = 1.29 kg,

m2 = 0.3 kg,

x1 = -0.75 m,

x2 = -0.2 m,

g = 9.8 m/s².

Substituting the information yields:

F_r = F_g

-k*x1 + k*x2 = m1*g - m2*g.

Solving gives k = 29.4 N/m.

b) To find the unloaded length l:

l = x1 - (F_1/k).

Given:

m1 = 1.95kg, x1 = -0.75m.

Substituting in provides:

l = x1 - (F_1/k) = 0.1.

3 0
2 months ago
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