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Anarel [89]
2 years ago
9

Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea’s reference frame is

13 km/h. The speed of the soccer ball from Sean’s reference frame is 22 km/h. Which conclusion is best supported by the information? A. Sean is standing still, and Rhea is running toward Sean while kicking the ball. B. Rhea is standing still after kicking the ball, and Sean is running away from Rhea. C. Both soccer players are standing still after the ball is kicked. D. Both soccer players are running while the ball is in motion.
Physics
1 answer:
BartSMP [9]2 years ago
5 0

Answer:  Sean is standing still, and Rhea is running toward Sean while   kicking the ball

Explanation: Your welcome :)

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In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component
Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0
2 years ago
The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me
uranmaximum [27]

Since the tower base is square with a side length of  125 m,

Therefore,

(125\ m)^2+ (125\ m)^2=31250 m^2

Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.

The volume of cylinder,

=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3

Thus, the mass of the air in the cylinder,

=1.225\ kg/m^3 \times 7948168.803\ m^3=9736506.78\ kg

Hence, the mass of the air in the cylinder is this more  than the mass of the tower.


4 0
1 year ago
An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the
laila [671]

Answer:

a)  V = 1.866 10² V ,  b)   V = 3.424 10⁵ V , c)   v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

           v = 0.027 c

           v = 0.027 3 10⁸

           v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

           Em₀ = U = q V

final point. Electrons with maximum speed

          Em_f = K = ½ m v2

          Em₀ = Em_{f}

          e V = ½ m v²

          V = ½ m v² / e

let's calculate

          V = ½  9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

          V = 1.866 10² V

           V = 1866 V

         

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

          V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

           V = 3.424 10⁵ V

           V = 342402 V

c)   this potential difference should give the protons the same speed as the electrons

             v = 8.1 10⁶ m / s

5 0
2 years ago
Un niño de 25 kg corre por un jardín con una velocidad de 2.5 m/s de forma que su trayectoria es tangente al borde de un carruse
Schach [20]

Answer:

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

Explanation:

Asumamos que tanto el niño como el carrusel no tienen carga externa aplicada sobre aquellos, de modo que se puede aplicar el Principio de Conservación de la Cantidad de Movimiento Angular:

m\cdot v \cdot R = (m\cdot R^{2}+I)\cdot \omega (1)

Donde:

m - Masa del niño, medida en kilogramos.

v - Velocidad lineal inicial del niño, medida en metros por segundo.

R - Radio máximo del carrusel, medida en metros.

I - Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.

\omega - Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.

Si sabemos que m = 25\,kg, v = 2.5\,\frac{m}{s}, R = 2\,m y I = 500\,kg\cdot m^{2}, tenemos que la velocidad angular final es:

\omega = \frac{m\cdot v\cdot R}{m\cdot R^{2}+I}

\omega = \frac{(25\,kg)\cdot \left(2.5\,\frac{m}{s} \right)\cdot (2\,m)}{(25\,kg)\cdot (2\,m)^{2}+500\,kg\cdot m^{2}}

\omega = 0.208\,\frac{rad}{s}

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

4 0
2 years ago
Scientific knowledge builds upon previous knowledge."
notka56 [123]

Answer:

B

Explanation:

Magic

4 0
2 years ago
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