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garik1379 [7]
2 years ago
8

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in

the process, the electron acquires a speed ve, while the proton acquires a speed vp. find the ratio ve/vp.
Physics
1 answer:
Yakvenalex [24]2 years ago
4 0
The charges are the same in absolute value, so the change of potential energy is the same. That means that the change in kinetic energy is also the same. Then:

1 = Ke/Kp = m_e *v_e^2 / m_p * v_p^2, or

v_e/v_p = sqrt( m_p/m_e),

So the speed of the electron will be sqrt( m_p/m_e) times greater than the speed of the proton
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What is the magnitude of the relative angle φ
melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0
1 year ago
Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo
timofeeve [1]

Answer:

a = 0.5 m/s²

Explanation:

Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:

ωf = ω₀ + α*t

⇒ ωf = α*t ⇒ α = \frac{wf}{t} = \frac{1.4 rad/s}{21 s} = 0.067 rad/s2

The angular velocity, and the linear speed, are related by the following expression:

v = ω*r

Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²

3 0
2 years ago
Romeo lanza suavemente guijarros a la ventana de julieta y quiere que los guijarros golpeen la ventana solo con con un component
Yuki888 [10]

Answer:

5.219\,\frac{m}{s}

Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)

v_{o,y} = 9.395\,\frac{m}{s}

0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t

\Delta t = 0.958\,s

Además, se determina el componente horizontal de la rapidez inicial (Likewise, the horizontal component of the initial speed is determined):

v_{o,x} = \frac{5\,m}{0.958\,s}

v_{o,x} = 5.219\,\frac{m}{s}

El guijarro tiene una rapidez de 5.219\,\frac{m}{s} cuando golpea la ventana (The peeble has a speed of  5.219\,\frac{m}{s} when it hits the window).

6 0
2 years ago
A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient
Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

5 0
2 years ago
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