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4 days ago

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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

are mA = 363 kg, mB = 517 kg, and mC = 154 kg. Find the magnitude and direction of the net gravitational force acting on each of the three particles (let the direction to the right be positive).Particle A is 0.5m from B and B is .25m from C... All in a straight line.

1 answer:

Sav [2.2K]4 days ago

3 0

The masses of particle A, B, and C are given, with all three particles aligned linearly. The distances between them are noted. The gravitational forces are attractive, compounding when acting in the same direction. The effects on each particle are formulated based on their distances.

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The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

inna [2205]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The velocity of the center of mass for the two-block system just prior to their collision is 2.9489 m/s

Explanation:

Provided information:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = collision time = 0.5 s

Question: What is the velocity of the center of mass for the two-block system right before the blocks collide, vf =?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It’s essential to calculate the required force:

$F=(m+m)*g*sin\theta$

Here, g = acceleration due to gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

6 0

20 days ago

Select the areas that would receive snowfall because of the lake effect.

Keith_Richards [2248]

Result:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

8 0

6 days ago

A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise

Ostrovityanka [2198]

Answer:

H = 109.14 cm

Explanation:

Given,

Assume that the total energy equals 1 unit.

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H

H = 0.475 x h

H = 0.475 x 2.3 m

H = 1.0914 m

H = 109.14 cm

6 0

15 days ago

If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t

Sav [2217]

Para calcular el peso utilizamos la fórmula:
$F=m*g=882N$

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
$F=G* \frac{ M_{p}*M_{E} }{ r^{2} }$
Donde:
G = constante de gravedad = $6.67* 10^{-11} m^{3} kg^{-1} s^{-2}$
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = $5.97* 10^{24} kg$
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
$m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }$

Resolviendo esta ecuación para r:
$r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}$
r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

3 0

22 days ago

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 [2593]

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

17 days ago