A top-fuel dragster accelerates from rest to a velocity of 100 m/s in 8 s. What is the acceleration?

An object has a mass of 5.20g. The mass of the object in kg is

vampirchik [111]

Answer:

0.0520kg

Explanation:

divide by 1000

8 0

3 years ago

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the

JulsSmile [24]

Answer:

(a) 3.9cm

(b) 1.66 x 10⁻⁸s

Explanation:

Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,

F = m x a --------------(i)

Where;

m = mass of the particle

a = acceleration of the mass

The centripetal acceleration is given by;

a = v² / r [v = linear velocity of particle, r = radius of circular path]

<em>Therefore, equation (i) becomes;</em>

F = m v²/ r --------------------(ii)

The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;

F = qvBsinθ -------------(iii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = angle between the velocity and the magnetic field

<em>Combine equations (ii) and (iii) as follows;</em>

m (v² / r) = qvBsinθ [divide both side by v]

m v / r = qBsinθ [make r subject of the formula]

r = (m v) / (qBsinθ) ---------(iv)

(a) From the question;

v = 1.48 x 10⁷m/s

B = 2.14mT = 2.14 x 10⁻³T

θ = 90° [since the direction of velocity is perpendicular to magnetic field]

m = mass of electron = 9.11 x 10⁻³¹kg

q = charge of electron = 1.6 x 10⁻¹⁹C

Substitute these values into equation (iv) as follows;

r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

r = 3.9 x 10⁻²m

r = 3.9cm

Therefore, the radius of the circular path is 3.9cm

(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

T = d / v --------------(*)

Where;

d = distance traveled in the circular path in one complete turn = 2πr

v = velocity of the motion = 1.48 x 10⁷m/s

d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]

d = 2(3.142)(3.9 x 10⁻²) = 0.245m

Substitute the values of d and v into equation (*) as follows;

T = 0.245 / 1.48 x 10⁷

T = 0.166 x 10⁻⁷s

T = 1.66 x 10⁻⁸s

Therefore, the time interval is 1.66 x 10⁻⁸s

6 0

3 years ago

during convection, hot material _____ (expands & sinks, cools & rises, expands & rises, deflates & rises) then m

Lady bird [3.3K]

Answer:

During convection, hot material expands & rises then moves to the side and cools & sinks. this circular pattern is called a convection current.

Explanation:

Convection is one of the three methods of transfer of heat. It occurs only in fluids (liquids or gases).

Convection occurs when there is a source of heat that heats a fluid, such as in a boiling pot of water. The water which is on the bottom of the pot becomes warmer before than the water at the top (because it is closer to the flame), and so it becomes less dense: for this reason, it expands and it becomes rising. On the contrary, the water on top is colder, so it is more dense and starts sinking, replacing the warmer water. As the new part of water gets warmer, it starts rising, and so the process is continuously repeated. This circular current is called convection current.

4 0

3 years ago

Read 2 more answers

Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz

amid [387]

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

7 0

4 years ago

Scenario

Anvisha [2.4K]

Answer:

1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

0 = y₀ + 0 - ½ g t2

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 2650/9.8)

t = 23.26 s

this is the horizontal scrolling time

x = v₀ t

x = 366.67 23.26

x = 8527 m

the speed at the point of arrival is

v_y = v_{oy} - g t = 0 - gt

v_y = - 9.8 23.26

v_y = -227.95 m / s

Module and angle form

v = $\sqrt{v_x^2 + v_y^2}$

v = √(366.67² + 227.95²)

v = 431.75 m / s

θ = tan⁻¹ (v_y / vₓ)

θ = tan⁻¹ (227.95 / 366.67)

θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

cos θ = v₀ₓ / v₀

sin θ = v_{oy} / v₀

v₀ₓ = v₀ cos θ

v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + vo sin θ t - ½ g t²

0 = -50 + vo sin 50 t - ½ 9.8 t²

x = x₀ + v₀ₓ t

0 = x₀ + vo cos θ t

0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

50 = 0,766 vo t – 4,9 t²

400 = 0,643 vo t

resolved

50 = 0,766 ( $\frac{400}{0.643 \ t}$ ) t – 4,9 t²

50 = 476,52 t – 4,9 t²

t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

t = [97.25 ± $\sqrt{97.25^2 - 4 \ 10.2}$ ] / 2

t = 97.25 ±97.04] 2

t₁ = 97.145 s

t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

t = 97.145 s

let's find the initial velocity

x = x₀ + v₀ cos 50 t

0 = -400 + v₀ cos 50 97.145

v₀ = 400 / 62.44

v₀ = 6.4 m / s

5 0

3 years ago