Which best describes the way a sound wave is sent trough the radio?

the engine of a car of amass of 2000 kg produced a force of 15000N find the acceleration of the car

weqwewe [10]

I believe the answer would be 7.5 m/s^2

3 0

2 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

2 years ago

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

olga_2 [115]

A) the final velocity = 66/9 m/s.
b) The total momentum before and after collision is the same because energy is destroyed or made.
Thanks brainly. <span />

7 0

3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

2 years ago

Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl

Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = ( $\frac{(m₂-m₁)}({m₁+m₂} * g$ = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0

2 years ago