All categories

2 years ago

Help with 3 please! I will give brainliest!

Physics

1 answer:

rusak2 [61]2 years ago

8 0

The answer is point b because vertical velocity is zero at the maximum height

You might be interested in

What happens to parallel light rays that strike a concave lens?

zhenek [66]

Due to the shape of the lens , parallel rays will be deviated

5 0

2 years ago

Read 2 more answers

Selena drew a diagram to show how current moves in a loop of wire that is placed between two magnets. At top left a piece of mag

OLEGan [10]

Answer:

B on Edge 2020

She can change the arrows so they show current traveling in opposite directions on the sides of the loop.

Explanation:

Just took the test haha

5 0

2 years ago

Read 2 more answers

A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.

harina [27]

I can not solve the problem if I do not have the mass.

3 0

2 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

2 years ago

Statistical time division multiplexing does not require the capacity of the circuit to be equal to the sum of the combined circu

aleksklad [387]

Answer:

The answer is True

Explanation:

Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.

In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.

Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.

7 0

2 years ago