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1 month ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

1 answer:

5 0

The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

serg [3582]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: The speed of a wave on a string under tension can be determined using the following:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

Distance a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5× $10^{-6}$ m.

When tension is doubled:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Distance in the same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With the increased tension, it moves $\Delta x =$ 15.4× $10^{-5}$ m

4 0

2 months ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Softa [3030]

The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "Doppler's effect."

The standard formula for the Doppler effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the observer relative to the medium =?.
$v_{s}$ = Velocity of the source in relation to the medium = 0 m/s.
$f_{o}$ = Frequency emitted from the source = 400 Hz.
$f$ = Frequency recognized by the observer = 408 Hz.

Substituting the given values into equation (A) will yield:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving the above will result in,
$v_{r}$ = 6.8 m/s

The correct result = 6.8m/s

7 0

2 months ago

Experimental tests have shown that hammerhead sharks can detect magnetic fields. In one such test, 100 turns of wire were wrappe

Ostrovityanka [3204]

B = µo*N*I/2r

Thus, B = 4πx10^-7*150*1.6/2*3.5 = 4.31x10^-5T

7 0

29 days ago

. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be

Softa [3030]

A 40 kg child throws a 0.5 kg stone at a velocity of 5 m/s. To find the recoil, we apply the conservation of momentum formula: m1•v1 + m2•v2 = 0, where m1 is the mass of the child, and v1 is the child's recoil velocity. Applying the known values results in 40•v1 = -0.5 × 5, leading to v1 = -2.5 / 40, which simplifies to v1 = -0.0625 m/s. Thus, the child's recoil speed is 0.0625 m/s.

6 0

1 month ago

Two charges of 15 pC and −40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through

Keith_Richards [3271]

To address this issue, we will utilize the principles related to Gauss' law, which states that the electric flux across a surface corresponds to the object's charge divided by the permittivity of vacuum. In mathematical terms, this can be expressed as

$\phi = \frac{Q_{net}}{\epsilon_0}$

It's crucial to remember that the net charge equals the difference between the two specified charges, so upon substitution,

$\phi = \frac{(15-40)*10^{-12}C}{8.85*10^{-12}C^2/Nm^2}$

$\phi = 2.82WB$

The negative sign indicates that the flux is directed into the surface

4 0

2 months ago