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Bad White [126]
4 years ago
14

How is speed independent of mass?

Physics
1 answer:
Yakvenalex [24]4 years ago
5 0

Answer:

The relationship is independent of the mass of the object escaping the massive body. Conversely, a body that falls under the force of gravitational attraction of mass M, from infinity, starting with zero velocity, will strike the massive object with a velocity equal to its escape velocity given by the same formula.

HOPE THIS HELPED!!!!!!!!!!!!!! XDDDDDDD

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A car is traveling north at 17.7m/s After 12 s its velocity is 14.1m/s in the same direction. Find the magnitude and direction o
dem82 [27]

Answer:

Option B

0.3 m/s2 South

Explanation:

Acceleration, a=\frac {v-u}{t} where v and u are final and initial velocities respectively, t is the time taken

Substituting 14.1 m/s for v, 17.7 m/s for u and 12 s for t then

a=\frac {14.1 m/s- 17.7 m/s}{12}=-0.3 m/s^{2}

Since this is negative acceleration, it's direction is opposite hence 0.3 m/s2 South

7 0
3 years ago
Nickel has a face-centered cubic structure and has a density of 8.90 g/cm3. what is its atomic radius?
vitfil [10]
Density = (mass/volume) 

<span>Mass = # atoms (in unit cell) x (mol / 6.022 x 10^23 atoms) x (58.69 g/mol) </span>
<span>The last number is the atomic mass of nickel </span>

<span>The number of atoms = 8(1/8) + 6(1/2) = 4 

</span>The volume (of the entire unit cell) = side^3 = (r x 8^(1/2))^3 

<span>6.84 g/cm3 = [4 atoms x (mol / 6.022 x 10^23 atoms) x (58.69 g/mol)] / [r x 8^(1/2)]^3 
</span>
<span> r = 1.36 x 10^-8 cm</span>
8 0
3 years ago
Read 2 more answers
Un campo eléctrico es creado por una carga eléctrica de 5 microcoulomb, a 35 cm se coloca una carga de prueba la cual es atraída
Sidana [21]

Answer:

Supongo que queremos determinar la carga de la carga de prueba.

Sabemos que la fuerza culombiana entre dos cargas q₁ y q₂, separadas por una distancia R, está dada por:

F = k_c*\frac{|q_1*q_2|}{R^2}

Aislandolo para una de las cargas, obtenemos:

\frac{F*R^2}{kc*|q_1|} =|q_2|

En este caso sabemos:

fuerza atractiva, por lo que los signos de las cargas son opuestos.

q₁  = 5 mC

R = 35cm

F = 1.5 N

Kc = 9*10^9 N*m^2/C^2

Un primer paso, seria reescribir todos los valores en las mismas unidades.

Sabiendo que:

100cm = 1m

R = 35cm = (35/100) m = 0.35m

Y sabiendo que:

1mc = 1*10^(-6) C

Entonces:

q₁  = 5 mC = 5*1*10^(-6) C = 5*10^(-6) C

Ahora podemos reemplazar esos valores en la ecuación de la fuerza, para obtener el valor de la otra carga:

\frac{15N*(0.35m)^2}{(9*10^9 N*m^2/C^2)*5*10^{-6}C} =|q_2| = 4.08*10^{-5} C

Y recordar que las cargas tienen signo opuesto, entonces la carga de la carga de prueba es:

q₂ = -4.08*10^-5 C

8 0
3 years ago
A pumpkin with a mass of 3.3 kg is launched from a catapult at an initial height of 3.8 m off the ground, with an initial speed
dimaraw [331]

Answer:

11.2m

Explanation:

Suppose the pumpkin is launched vertically and the speed of 13.9 m/s is absolutely vertical. As the pumpkin rises up, its kinetic energy is converted to potential energy, let the reference point be the round, we can create the following equation from the law of energy conservation:

E_1 + P_1 = E_2 + P_2

0.5mv_1^2 + mgh_1 = 0.5mv_2^2 + mgh_2

we can substitute g = 9.8m/s^2, v_1 = 13.9m/s, v_2 = 0.5v_1 = 0.5*13.9 = 6.95 m/s, h_1 = 3.8

We can also divide both sides by m and 0.5

13.9^2 + 2*9.8*3.8 = 6.95^2 + 2*9.8*h_2

19.6h_2 = 219.3875

h_2 = 11.2m

5 0
4 years ago
Which model correctly represents the product or products of the chemical reaction?
fgiga [73]

Answer:

you can go to quizlet in really helps.

Explanation:

7 0
3 years ago
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