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3 years ago

14

Three of the forces in nature attract particles together, with varying strengths, to build the universe. Which is the only one t

hat energizes the universe by making nuclear fusion possible

1 answer:

EastWind [94]3 years ago

3 0

Answer:

The photon or quantum of light

Explanation:

You might be interested in

Name on contact force and one field force acting on you right now

kati45 [8]

Answer:

Contact forces are forces that require the actual contact (touching) of two pieces of matter. There are a variety of contact forces. A very common one is friction. Anytime that two surfaces are in contact with one another, there is friction between the two surfaces. A field force is a force that works at a distance. No touching is required. Gravity is a good example of a field force, because it works whether or not an object is touching something or touching nothing at all.

8 0

3 years ago

) A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward

Talja [164]

Answer:

The time interval during which the rocket engine provides upward acceleration is 2.1 s

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine burnout:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial velocity

t = time

a = upward acceleration

g = acceleration due to gravity (downward)

The velocity of the rockey is given by this equation:

v = v0 + a · t (v0 = 0 because the rocket is launched from rest)

v = a · t

and after burnout:

v = v0 + g · t

Where v = velocity at time t

We know that when the altitude is 64 m the velocity is 60 m/s. Then let´s use the following equation system:

y = y0 + v0 · t + 1/2 · a · t² (y0 and v0 = 0)

v = a · t

Then:

64 m = 1/2 · a · t²

60 m/s = a · t

a = 60 m/s / t

Replacing "a = 60m/s / t" in the equation of height:

64 m = 1/2 ·( 60m/s / t) · t²

64 m = 30 m/s · t

t = 64 m / 30 m/s

t = 2.1 s

Then, the time interval during which the rocket engine provides upward acceleration is 2.1 s

5 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0

3 years ago

What is the period of a wave that travels through a spring at 2.5 m/s and has a wavelength of 3m ?

Bond [772]

Answer:

Explanation:

f

=

v

λ

where v is the velocity of the wave

λ

is the wavelength.

3 0

3 years ago

A sinusoidal wave has the following wave function: y(x,t) = (2.5 m) sin[(3.0 m-1) x – (24 s-1) t + π/2] What is the frequency of

Vera_Pavlovna [14]

Answer:

3.82 Hertz

Explanation:

$y = 2.5 Sin\left (3x-24t+\frac{\pi }{2} \right )$

This is the equation of a wave which varies sinusoidally.

The standard equation of a wave is given by

$y = A Sin\left ( kx-\omega t+\phi \right )$

where, A be the amplitude of the wave, k be the wave number, x be the displacement of wave, ω be the angular frequency and t be the time taken, and Ф be the phase angle.

now compare the given equation by the standard equation, we get

k = 3

ω = 24

Ф = π / 2

So, the angular frequency = 24

The relation between the angular frequency and the frequency is given by

ω = 2 x π x f

24 = 2 x 3.14 x f

f = 3.82 Hertz

4 0

3 years ago