All categories

3 years ago

Physics HELP!

Physics

1 answer:

shusha [124]3 years ago

6 0

Answer:

1. B

2. A

Explanation:

1. The displacement is the change in position. At t = 0, x = 1.0. At t = 8.0, x = 6.0. So from t=0 to t=8, Δx = 6.0 − 1.0 = 5.0.

2. The instantaneous velocity is the slope of the tangent line at any point of a position vs. time graph.

The average velocity is the displacement divided by the time interval.

You might be interested in

Which property of light remains unchanged when it enters a different medium?

Amiraneli [1.4K]

The FREQUENCY of light remains unchanged once it leaves the source.

3 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Natasha_Volkova [10]

Ok srry bout last time but the answer is A) kelvin: time the reason is because everything else is a SI. Hope this helps and TURTLE

7 0

3 years ago

Read 2 more answers

what is one example of the use of an infrared sensor? Track shipments, monitor building exit and entry, water flows without turn

kap26 [50]

Your answer should be water flows without turning on a facet. Hope this helps!

6 0

3 years ago

Read 2 more answers

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$

There is NO initial vertical velocity, so:

$\large\boxed{\Delta d= \frac{1}{2}at^2}}$

Rearrange to solve for time:

$2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}$

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

$t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s$

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

$\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}$

6 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

3 years ago