Which pair of properties apply to both sound and electromagnetic waves?

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for

emmasim [6.3K]

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

4 0

3 years ago

While visiting the beach, you enjoy the warm ocean water, but the sand burns your feet. That night you walk along the beach and

garik1379 [7]

The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

8 0

3 years ago

PLEASE HELP!! ILL MARK BRAINLYEST!!

topjm [15]

Answer: True

Explanation:

Mass is the amount of matter in an object. Everything is made up of matter.

3 0

3 years ago

Read 2 more answers

La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s

NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

For the waves in this string we have:

$f=625 Hz$ , since it completes 625 cycles per second

$\lambda=3.33 cm = 0.033 m$ is the wavelength

So the speed of the wave is

$v=(625)(0.0333)=20.6 m/s$

The speed of the waves in a string is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$ (1)

where

T is the tension in the string

$\mu=\frac{m}{L}$ is the linear density

In this problem:

$m=16.5 g = 16.5\cdot 10^{-3} kg$ is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

$T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N$

8 0

3 years ago

A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force

MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = $\frac{D}{2}$

= $\frac{200}{2}\ m$

= 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

Force F = $\frac{mv^2}{R}$

= $\frac{1500\times (25)^2}{100}\ N$

= 9375 N. [ANS}

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.

6 0

3 years ago