2. A car traveling 300 miles in 5 hours is an example of

What determines the quality of a conductor

aksik [14]

<span>number of free electrons present.</span>

8 0

3 years ago

Suppose that the distance an aircraft travels along a runway before takeoff is given by Upper D equals (5 divided by 3 )t square

Dahasolnce [82]

Answer:

a. Time=25seconds

b.distance=1041.67m

Explanation:

a.The equation for $D$ in terms of m/s is $\frac{250}{3}m/s$ after conversion.

To find when speed reaches 300km/hr=83.33m/s, we find $D\prime$ and solve for $t$

$D=\frac{5}{3}t^2\\D\prime=\frac{5}{3}(2t)=\frac{10}{3}t=\frac{250}{3}\\t=25sec$

b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.

#To find distance travelled in that time , we substitute for $t=25$ in our distance equation:

$D(25)=\frac{5}{3}(25)^2\\=1041.67m$

Hence the distance of the plane before it gets airborne is 1041.67m

4 0

3 years ago

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

$d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m$

Next, you calculate the wavelength of the sound waves by using the following formula:

$\lambda=\frac{v_s}{f}$

vs: speed of sound = 343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

$\lambda=\frac{343m/s}{400Hz}=0.8575m$

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

$\Delta d=3.827m-2.10m=1.727m$

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

$\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2$

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0

2 years ago

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

Yuri [45]

Answer:

$q = 2.067 \times 10^{-5}\ C$

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

$q =\dfrac{mg}{E}$

$q =\dfrac{0.00141 \times 9.81}{670}$

$q = 2.067 \times 10^{-5}\ C$

Charge of the particle is equal to $q = 2.067 \times 10^{-5}\ C$

6 0

3 years ago

Amount of force exerted on an object due to gravity is called

MAXImum [283]

The gravitational force between a mass and the Earth is the object'sweight. Mass is considered a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: Fg = mg.

Hoped this helped!

6 0

3 years ago