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3 years ago

Which of the following are true for an object in projectile motion?

1 answer:

4 0

The correct options are: B, C and D.

In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with $g=9.81 m/s^2$ (acceleration of gravity)
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.

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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

3 years ago

Two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of

olga2289 [7]

Answer:

c. because A will land first becuase its heavier :)

Explanation:

8 0

3 years ago

What are forces that two objects apply on each other

liq [111]

Answer: Whenever two objects are touching, they usually exert forces on each other. The force of gravity, on the other hand, is an example of a force that exists between objects without them having to be in contact. Objects with mass exert forces on each other via the force of gravity.

HOPE IT HELPED:) HAVE A NICE DAY

3 0

3 years ago

Vector question,university physics zemanski

just olya [345]

Answer:

yes This is correct Answer

3 0

3 years ago