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6 days ago

When Royce was 10 years old, he had a mass of

1 answer:

6 0

The gravitational force between Royce and Earth would double by the age of 16. Newton’s law of universal gravitation states that the gravitational force is proportional to the masses involved and inversely proportional to the square of the distance between them. At age 10, Royce weighed 30 kg, and by age 16, he weighed 60 kg. Since his mass doubled from 10 to 16 years, this results in a corresponding doubling of the gravitational force.

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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [2668]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

1 month ago

Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be loca

Keith_Richards [2907]

Response:

1) An observer in B 'perceives the two events occurring at the same time

2) Observer B recognizes that the events happen at different times

3) Δt = Δt₀ /√ (1 + v²/c²)

Clarification:

This scenario illustrates the concept of simultaneity in special relativity. It is important to keep in mind that light's speed remains constant across all inertial frames

1) Since the events are stationary within the frame S ', they propagate at the constant speed of light, resulting in them reaching observation point B'—located equidistantly between both events—simultaneously

Thus, an observer in B 'observes the two events occurring at the same time

2) For an observer B situated within frame S attached to the Earth, both events at A and B appear to take place at the same moment. However, the event at A covers a shorter distance, while the event at B travels a longer distance, since frame S 'is in motion at velocity + v. Hence, with a constant speed, the event covering the lesser distance is perceived first.

Consequently, observer B perceives that the events do not occur simultaneously

3) Let's determine the timing for each event

Δt = Δt₀ /√ (1 + v²/c²)

where t₀ represents the time in the S' frame, which remains at rest for the events

8 0

18 days ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

ValentinkaMS [3091]

No one is going to handle that for a mere 5 points lol.

8 0

1 month ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [2968]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

1 month 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 [2826]

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

1 month ago