in which type of wave are vibrations at right angles to the direction in which the wave is travelling

What do we mean by a penetrative pass in the game of football?

Ivanshal [37]

Answer:

Penetration means forward passes can go through the opposition lines. Once these penetrative passes get through each line, it eliminates the line of players it broke through and leaves the player in possession closer to the opposition goal.

Explanation:

6 0

3 years ago

A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce

Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

$x=x_0+v_0t+at^2$

Where

$x_0 =$ Initial position

$v_0 =$ Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

$v_f = v_0 + at$

Where,

$v_f = Final velocity$

For our case we have that there is neither initial position nor initial velocity, then

$x= at^2$

With our values we have $x = 401.4m, t=4.945s$ , rearranging to find a,

$a=\frac{x}{t^2}$

$a = \frac{ 401.4}{4.945^2}$

$a = 16.41m/s^2$

Therefore the final velocity would be

$v_f = v_0 + at$

$v_f = 0 + (16.41)(4.945)$

$v_f = 81.14m/s$

Therefore the final velocity is 81.14m/s

8 0

3 years ago

A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo

WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

$F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}$

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

$acceleration=\frac{30}{5}=6m/s^{2}$

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

$v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds$

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0

4 years ago

A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W

Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s$

Since, the motion is downward, final velocity must be negative. So,

$v_d=-4.95\ m/s$

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity ( $v_d$ ) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s$

Since, the motion is upward, final velocity must be positive. So,

$v_{up}=3.31\ m/s$

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

$J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns$

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

5 0

3 years ago

Some scientists look at the boundary between two biological communities and see a sharp dividing line. Others looking at the sam

Elena L [17]

This is for the reason that individuals are not continually taking a gander at precisely the same, and on the grounds that individuals' psyches of ten work distinctively and process data in marginally extraordinary ways getting diverse understandings of similar information.

7 0

3 years ago

in which type of wave are vibrations at right angles to the direction in which the wave is travelling​

in which type of wave are vibrations at right angles to the direction in which the wave is travelling