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My name is Ann [436]

2 years ago

11

A cheetah has an acceleration rate of 5.80 m/s2 . What is the cheetah’s final velocity after 7.80 seconds, if initially, his vel

ocity is 3.50 m/s?

1 answer:

Olegator [25]2 years ago

4 0

Answer:

you got holbrook huh

Explanation:

me too

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Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver

zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0

2 years ago

Suppose you have two identical sheets of paper and you crumple one of the sheets of paper into a ball. If you drop the crumpled

sammy [17]

Answer:

Because it can easily resist air resistance.

Explanation:

Since air resistance is not negligible, the crumpled paper will reach the ground first because it can easily resist air resistance surrounding it compare to the un-crumpled one that will be influenced by the air thereby causing the un-crumpled paper to spend more time in the air

3 0

2 years ago

When was the first airplane made?

expeople1 [14]

The first ariplanr was made December 17, 1903

3 0

3 years ago

Read 2 more answers

A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference

Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

$q=CV$

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

$C'=kC$

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

$q=C'V'$

So we can combine the two equations:

$CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}$

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

$V'=\frac{71.0}{4.50}=15.8 V$

6 0

2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago