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3 years ago

3. A simple way to state Newton’s first law is:

Physics

1 answer:

MrRissso [65]3 years ago

4 0

Explanation:

A simple way to state Newton's first law is:

For every action force, there is a reaction force which is equal in magnitude and opposite in direction.

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Two identical objects (A, B) travel circles of the same radius, but object A completes three times as many rotations as object B

malfutka [58]

Answer:

a) One-ninth the force acting on object A.

Explanation:

First, we derive an expression for the centripetal force acting on both objects.

For object A, centripetal force is:

$F_A = \frac{m{v_A}^2}{r}$

For object B, centripetal force is:

$F_B = \frac{m{v_B}^2}{r}$

We are given that they have the same mass and they move in circles of the same radius.

If object A completes three times as many rotations as object B, then, object must have 3 times the speed of object B.

Hence:

${v_A} = 3*{v_B}$

Therefore, $F_A$ becomes:

$F_A = \frac{m({3*v_B}^{2} )}{r}\\\\\\F_A = \frac{9m{v_B}^{2}}{r}$

$F_A = 9F_B$

=> $F_B = \frac{1}{9} F_A$

Therefore, the net centripetal force acting on object B is one-ninth of the force acting on object A.

4 0

3 years ago

5. A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 6400 m away?

EastWind [94]

Answer:

8.89 secs

Explanation:

720m = 1 sec

6400m = ?

6400/720 = 8.89 secs

8 0

3 years ago

A constant horizontal F force began to act on the initially immovable body placed on a horizontal surface. After t time the forc

mel-nik [20]

Answer:

The coefficient of friction is (F/(19.6·m)

Explanation:

The given parameters are;

The force applied to the immovable body = F

The time duration the force acts = t

The time the body spends in motion = 3·t

The acceleration due to gravity, g = 9.8 m/s²

From Newton's second law of motion, we have;

The impulse of the force = F × t = m × Δv₁

Where;

Δv₁ = v₁ - 0 = v₁

The impulse applied by the force of friction, $F_f$ is $F_f$ × (3·t - t) = $F_f$ × (2·t)

Given that the motion of the object is stopped by the frictional force, we have;

The impulse due to the frictional force = Momentum change = m × Δv₂ = $F_f$ × (2·t)

Where;

Δv₂ = v₂ - 0 = v₂

Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to act, v₁, we have;

m × Δv₂ = $F_f$ × (2·t) = m × Δv₁ = F × t

∴ $F_f$ × (2·t) = F × t

$F_f$ = F × t/(2·t) = F/2

The coefficient of dynamic friction, $\mu _k$ = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)

$\mu _k$ = (F/(19.6·m)

The coefficient of friction, $\mu _k$ = (F/(19.6·m)

5 0

3 years ago

How is this formula ( Rt= r1 x r2 / r1 + r2) which is used to calculate total resistance of two parallel resistors found from th

Maru [420]

Explanation:

The total resistance used to calculate the total resistance of the two parallel resistor is given by :

$R_t=\dfrac{r_1\times r_2}{r_1+r_2}$

Taking reciprocal of the above equation.

$\dfrac{1}{R_t}=\dfrac{r_1+r_2}{r_1\times r_2}$

We can also write the above equation as :

$\dfrac{1}{R_t}=\dfrac{r_1}{r_1\times r_2}+\dfrac{r_2}{r_1\times r_2}$

On simplification of above equation,

$\dfrac{1}{R_t}=\dfrac{1}{r_2}+\dfrac{1}{r_1}$

$\dfrac{1}{R_t}=\dfrac{1}{r_1}+\dfrac{1}{r_2}$

Hence, proved.

4 0

3 years ago

A clindrical rod of uniform density is located with its center at the origin, and its axis along the axis. It rotates about its

zhenek [66]

Answer:

6093.2328 J

Explanation:

For cylindrical rod moment of inertia will be

$I_X=I_Y=\frac{1}{12}m(3r^2+h^2)$

$I_X=I_Y=\frac{1}{12}\times 4(3\times 0.06^2+0.6^2)=0.1236$ $kg -m^2$

we have given time =0.02 sec

Angular speed = $\frac{2\pi }{T}=\frac{2\times 3.14}{0.02}=314\ rad/sec$

Rotational KE = $\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.1236\times 314^2=6093.2328\ J$

5 0

4 years ago