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3 years ago

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A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

esistance) the ball will land on the ground with a speed of

1 answer:

NikAS [45]3 years ago

5 0

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

$KE_i + U_i = KE_f + U_f$

here we know that

$Ui = U_f = 0$ (as it will be on ground at initial and final position)

so we will say

$KE_i = KE_f$

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

$v_f = v_i = 30 m/s$

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A charge enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc

frozen [14]

Answer:

The new radius is three times the initial radius

Explanation:

The magnetic force is given by the equation

F = q v x B

Where the blacks indicate vectors, q is the electric charge, v the velocity of the particle and B the magnetic field. The scalar magnitude of this force is

F = q v B sin θ

Where θ is the angle between the velocity and the magnetic field.

As the force is perpendicular to the velocity, the particle describes a circular motion, using Newton's second law we can find the acceleration that is centripetal.

F = ma

a = v² / r

q v B = mv² / r

R = mv / qB

Let's calculate for the new speed (v₂ = 3v)

R₂ = (m / qB) 3v

R₂ = 3 R

R₂ / R = 3

The new radius is three times the initial radius

7 0

3 years ago

A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have?

vaieri [72.5K]

Answer:

30000 J = 30 kJ

Explanation:

Kinetic energy is the energy possessed by a moving object solely due to its motion.

You can get the K.E. of an object using the equation,

K.E. = $\frac{1}{2} mv^{2}$

So you get,

K.E. = $\frac{1}{2} 600×10^{2}$

= 30000 J = 30 kJ

3 0

2 years ago

The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.

ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

$\lambda=\frac{h}{p}$

where

$h=6.6 \cdot 10^{-34} Js$ is the Planck constant

p is the momentum of the particle

In this problem, the momentum of the electron is equal to the product between its mass and its speed:

$p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s$

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

$\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m$

So, the answer is True.

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2 years ago

What is the force on an object that goes from 35 m/s to 85 m/s in 20 seconds and has a mass of 148 kg

Sever21 [200]

F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N

3 0

3 years ago

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

6 0

2 years ago