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abruzzese [7]
3 years ago
8

Newton's third law states that forces always occur in

Physics
2 answers:
Arada [10]3 years ago
7 0
Answer:
C enqual, opposite
nika2105 [10]3 years ago
6 0

Answer:

Equal , paralell

Explanation:

You might be interested in
-<br> ZOOLS<br> 6) The mass of a motorcycle is 250 kg. What is?<br> A) Its weight on Earth in Newtons?<br> B) Its weight on the
marishachu [46]

Answer:

Explanation:

Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.

A)

F = m * a

m = 250 kg

a = 9.81 m/s^2

F = 250 * 9.81 = 2452.5 N

B)

The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.

a = 9.81 + (1/6) = 1.635

m = 250

F = 250 * 1.635

F = 408.75

C)

The mass is the same anywhere in the universe.

250 kg

4 0
3 years ago
A concave mirror is used to ______ ( converge, diverge) light and a convex mirror is used to _____( diverge or converge )
Colt1911 [192]

Answer:

converge, diverge

Explanation:

4 0
3 years ago
Read 2 more answers
The container was lifted a height of 14 m
Alika [10]
Work done = Force X Distance

3 430 000J = Force X 14m

Force = 3 430 000J / 14m
= 245 000 N

Hope this helps!
8 0
2 years ago
Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido
almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

v_{P/B} = v_{P} - v_{B} (1)

Donde:

v_{P/B} - Velocidad del perro relativo al barco, en kilómetros por hora.

v_{P} - Velocidad del perro con respecto al observador, en kilómetros por hora.

v_{B} - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que v_{B} = 30\,\frac{km}{h} y v_{P/B} = 10\,\frac{km}{h}, entonces la velocidad del perro con respecto al observador es:

v_{P} = v_{B} + v_{P/B}

v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}

v_{P} = 40\,\frac{km}{h}

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0
3 years ago
A merry-go-round accelerating uniformly from rest achieves itsoperating speed of 2.5rpm in five revolution.
g100num [7]

Answer:

(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²

(B) Time of motion 240.2 s

Explanation:

Given;

final angular speed, ωf = 2.5 RPM

angular distance, θ = 5 rev = 5 x 2π = 10π rad

initial angular speed, ωi = 0

final angular speed in rad/s;

\omega_f = \frac{2.5 \ rev}{min} \times \ \frac{2\pi}{1 \ rev} \times \ \frac{1 \min}{60 s} = 0.2618 \ rad/s \\

(A) the magnitude of its angular acceleration(rad/s^2);

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2\times 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = \frac{0.0685}{20\pi} \\\\\alpha = 1.09\times 10^{-3} \ rad/s^2

(B) Time of motion;

\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09\times 10^{-3} t\\\\t = \frac{0.2618}{1.09\times 10^{-3}} \\\\t = 240.2 \ s

7 0
3 years ago
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