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3 years ago

A rocket experiences a 45,000 N force as it accelerates at 30 m/s/s. What is the mass of the rocket?

Physics

1 answer:

Gre4nikov [31]3 years ago

6 0

Answer:

15,00kg

Explanation:

here's your solution

=> force = 45,000N

=> acceleration = 30m/s^2

=> mass = ?

=> Mass = force/acceleration

=> mass = 45,000/30

=> mass = 15,00kg

hope it helps

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A 0.780 m long section of cable carrying current to a car starter motor makes an angle of 60.0° with the Earth's 5.50 ✕ 10−5 T f

STatiana [176]

Answer:

188 A

Explanation:

Parameters given:

Length of cable, L = 0.78 m

Angle, θ = 60º

Magnetic field, B = 5.5 * 10^(-5) T

Force experienced by wire, F = 7 * 10^(-3) N

The force experienced by a current carrying wire of length L, due to a magnetic field B is given as:

F = I * L * B * sinθ

=> I = F/(L * B * sinθ)

I = (7 * 10^(-3)) / (0.78 * 5.5 * 10^(-5) * sin60)

I = 188 A

6 0

3 years ago

Read 2 more answers

.If the atoms that share electrons have an unequal attraction for the electrons, the bond is calleda. nonpolar. c. ionic.b. pola

Fittoniya [83]

Answer:

The right option is the 'polar' option.

Explanation:

- The right answer cannot be ionic as sharing of electrons is mentioned. Ionic bonds don't involve the sharing of electrons

- Polarity in terms of bonding is used to describe the situation when one of the atoms involved in the sharing of electrons is more electronegative than the other and pulls the shared electrons more towards itself, thereby inducing a slight positive charge on the other atom and gaining a slight negative charge itself.

- In the absence of this kind of scenario in a covalent bonding that sharing occurs, then it is described as non polar. The two atoms that participate in the sharing of electrons here have similar electronegativities and thereby leave the shared electrons at the middle.

- Dipolar compound is used to refer to organic molecules that are essentially neutral but carry a positive & negative charge in one of their various existence forms.

3 0

4 years ago

Un adolescente que va en monopatín rueda hacia abajo sobre un plano inclinado de 18.0 m de largo. El chico parte con una rapidez

qaws [65]

Answer: 31.62°

Explanation:

Tenemos como datos:

Distancia = 18.0m

Velocidad inicial = 2.0 m/s

Tiempo total = 3.3s

Sabemos que para un plano inclinado (ignorando el rozamiento) la aceleración se escribe como:

a(t) = g*sen(θ)

donde θ es el ángulo del plano inclinado, y g = 9.8m/s^2

Sabemos que para la velocidad tenemos que integrar la aceleración sobre el tiempo, entonces:

v(t) = g*sen(θ)*t + v0

Donde v0 es la velocidad inicial: v0 = 2.0m/s

v(t) = 9.8m/s^2*sen(θ)*t + 2.0m/s

Y para la posición, podemos integrar de vuelta sobre el tiempo:

p(t) = 0.5*9.8m/s^2*sen(θ)*t^2 + 2.0m/s*t + p0

Donde p0 es la posición inicial, podemos considerar que es cero para este problema.

p(t) = 4.9m/s^2*sen(θ)*t^2 + 2.0m/s*t

Y usando los datos iniciales, sabemos que en 3.3 segundos se recorren 18 metros, entonces:

p(3.3s) = 18m = 4.9m/s^2*sen(θ)*(3.3s)^2 + 2.0m/s*3.3s

18m = 51.744m*sen(θ) + 6.6m

sen(θ) = (18m - 6.6m)/ 51.744m

θ = cosec( (18m - 6.6m)/ 51.744m ) = 31.62°

4 0

3 years ago

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The larg

jeyben [28]

Answer:

<h2>Magnetic field strength in that region is 1.2 T</h2>

Explanation:

As we know by the formula of radius of charge moving in external field is given as

$R = \frac{mv}{qB}$

so we will have

$R = 25 cm$

$m = 1.6 \times 10^{-27} kg$

$q = 1.6 \times 10^{-19} C$

$v = 0.10 \times 3 \times 10^8 m/s$

now we have

$0.25 = \frac{(1.6 \times 10^{-27})(3\times 10^7)}{(1.6 \times 10^{-19})B}$

now we have

$B = 1.2 T$

8 0

3 years ago

Rutherford discovered the nucleus of the atom by firing a particles at gold foil. An a particle has a charge of q = +2e and a ma

Readme [11.4K]

Answer: $r_0=3.037\times 10^{-14}m$

Explanation:

Given

charge on alpha particle=+2e

mass of alpha particle= $6.64\times 10^{-27}$ kg

Charge on gold nucleus=+79e

Velocity at r=1m is $1.9\times 10^{7}$

Using Energy conservation

Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus

therefore

$\frac{1}{2}mv^2+U_{r=1m}=U_{closest\ to\ nucleus}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )+\frac{K\left ( 2e\right )\left ( 79e\right )}{1}=\frac{K\left ( 2e\right )\left ( 79e\right )}{r_0}$

$\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]$

on solving we get

$\frac{1}{r_0}=3.292\times 10^{13}$

$r_0=3.037\times 10^{-14}m$

8 0

3 years ago