All categories

3 years ago

Can someone spell the word hi for me I’ll give you 30 points

Physics

2 answers:

Fiesta28 [93]3 years ago

8 0

"Hi"

its spelled with an H & an I :^

jasenka [17]3 years ago

3 0

Answer:

hi poop

Explanation:

You might be interested in

HALP me!! This is a physics question.

emmasim [6.3K]

<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>

<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 = 50000 N

Distance = ?

Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

4 0

3 years ago

Read 2 more answers

A spider twirls a 25 mg fruit fly around in a circle with radius 17.6 cm at the end of a web. If the velocity of the fly is 110

olchik [2.2K]

Answer:

Fc = 1.7x10^-4 N

Explanation:

Convert everything to proper units:

m = 25mg = 2.5x10^-5 kg

r = 17.6cm = 0.176m

v = 110cm/s = 1.1m/s

the formula for centripetal force is Fc = mv^2 / r

Plug everything and solve for Fc;

fc = (2.5x10^-5)(1.1^2) / 0.176

Fc = 1.7x10^-4 N

8 0

3 years ago

A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th

Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

$Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

now we have

$W_{gravity} = -mg(Lsin\theta)$

$W_{friction} = -\mu mgcos(\theta) L$

so from above equation

$Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

so from above equation work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

8 0

3 years ago

Sean, after being so happy for two full days that he reported he "never needed much sleep," now is stating he is so sad that he

Kaylis [27]

D or A i'm not that sure

5 0

2 years ago

Read 2 more answers

Unpolarized light, with an intensity of I0, is incident on an ideal polarizer. A second ideal polarizer is immediately behind th

user100 [1]

Answer:

The light transmitted through the system will be 0.125*I₀.

Explanation:

The light transmitted through the system can be found using Malus Law:

$I = I_{0}cos^{2}(\theta)$ (1)

Where:

I: is the intensity of the light transmitted

I₀: is the initial intensity

θ is the angle relative to the first polarizer’s = 60°

Because the light transmitted by the first polarizer is dropped by half, the equation (1) results as:

$I = \frac{I_{0}}{2}cos^{2}(\theta)$

$I = \frac{I_{0}}{2}cos^{2}(60)$

$I = 0.125I_{0} = \frac{1}{8}I_{0}$

Therefore, the light transmitted through the system will be 0.125*I₀.

I hope it helps you!

7 0

2 years ago