Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. its average angular acceleration is

MaRussiya [10]

Answer:

He sowed the seeds

Explanation:

While the question should have given the person who did the sowing for example the seeds were sown by him/ her/ the farmer/ or any name. Therefore, voice is given in passive and to change passive voice to active voice then the sentence will read as follows assuming that the seeds were sown by him

He sowed the seeds

7 0

1 year ago

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$

F) -11.1 m/s

The y-component of the velocity at time t is given by:

$v_y(t) = v_y + at$

where

$v_y = 9.3 m/s$ is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

$v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s$

where the negative sign means the vertical velocity is now downward.

3 0

2 years ago

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to

mote1985 [20]

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

6 0

1 year ago

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

Korolek [52]

Answer:

a

$z= 2.5 \ m$

b

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

so

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

$z = \frac{5 + 3}{2}$

$z = 4\ m$

and

$z = \frac{ 5 -3 }{2}$

=> $z = 1 \ m$

=> $z = (1 \ m , 4 \ m )$

7 0

2 years ago

How many ternary strings of length 2 n are there in which the zeroes appear only in odd-numbered positions?

Rashid [163]

In a string of length $2n$ , there are exactly $n$ odd-numbered posiitons. In the remaning $n$ even-numbered positions, we want to choose from only two possible digits, either a 1 or a 2. This means there are $2^n$ such ternary strings.

8 0

2 years ago