Ask question

Ask question

All categories

Genrish500 [490]

2 years ago

5

For a particular type of motion, the velocity is zero but the speed is a nonzero quantity. Which statement can you make about th

is motion

1 answer:

masha68 [24]2 years ago

8 0

Answer:

because speed is the modulus of velocity which is a vector

the velocity to be zero it must be a round trip

Explanation:

This is because speed is the modulus of velocity which is a vector.

For the velocity to be zero it must be a round trip, therefore the resulting vector zero

On the other hand, the speed of the module is the same in both directions

You might be interested in

Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as

Sonja [21]

Explanation:

The waveform expression is given by :

$y(x,t)=y_m\ sin(0.333x+5.36+585t)$ ...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

$y(x,t)=y_m\ sin(kx+\phi+\omega t)$ ..........(2)

Where

$k=\dfrac{2\pi}{\lambda}$

$\omega=2\pi f$

On comparing equation (1) and (2) we get :

$0.333=\dfrac{2\pi}{\lambda}$

$\lambda=18.86\ m$

$585=2\pi f$

f = 93.10 Hz

Time period, $T=\dfrac{1}{f}$

$T=\dfrac{1}{0.010}$

T = 0.010 s

Phase constant, $\phi=5.36\ radian$

Hence, this is the required solution.

8 0

2 years ago

Current X is 2.5 A and runs for 39 seconds. Current Y is 3.8 A and runs for 24 seconds. Which current delivered more charge, and

Aleonysh [2.5K]

Answer: B. Current x delivered 6.3 C more then Y

Explanation:

7 0

2 years ago

Lidia plans an experimental investigation to see how the thickness of a lens affects the point where a beam of light is focused.

alexandr1967 [171]

Answer:

The type of light and the material of lenz.

Explanation:

1) As the investigation is based on how the thickness of a lens effect the other variable. Thickness of the lenz is independent variable. So Lidia has to experiment with the different thicknesses in order to find the effect on dependent variable.

2) As the investigation is based to find the point where the beam of light is focused. It's a dependent variable and Lidia has no control over it. So the only thing she can do is to measure and observe how it respond to the changes in independent variable.

3) For conclusion, she has to make sure that the other variables are not effecting the output or results that is the beam point where the light is focused. So she must have to kept constant the type of light and material of lenz otherwise she won't be able to discriminate the effect of thickness of lenz from other causes.

8 0

2 years ago

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0

2 years ago

A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0

2 years ago