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3 years ago

6

A soccer ball kicked on a level field has an initial vertical velocity component of 15.0 mps. assuming the ball lands at the sam

e height from which it was kicked was the total time of the ball in the air?

1 answer:

telo118 [61]3 years ago

5 0

Answer:

The time taken for the ball to fly up in the air and back down again is 3.058 seconds.

Explanation:

Since the ball ends up at the same vertical distance ( on the ground) as it was at the start of its motion, we can set the total displacement of the ball equal to 0.

Thus, this problem can be simply solved by the following equation of motion:

$s = u*t + \frac{1}{2} (a*t^2)$

Here, s = total change in distance = 0 m

u = initial speed = 15 m/s

a = acceleration due to gravity = -9.81 m/s^2

t = time (to be found)

Substituting these values in the equation we get:

$0=15t+0.5(-9.81t^2)$

$-15t = -4.905t^2$

$t=15/4.905$

t = 3.058 seconds

So, the time taken for the ball to fly up in the air and back down again is 3.058 seconds.

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light of a certain frequency has a wavelength of 438 nm in water.What is the wavelength of this light in benzene

sveticcg [70]

Answer:

388.97 nm

Explanation:

The computation of the wavelength of this light in benzene is shown below:

As we know that

n (water) = 1.333

n (benzene) = 1.501

$\lambda (water) \times n(water) = \lambda (benzene) \times n(benzene)$

And, the wavelength of water is 438 nm

$\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]$

Now placing these values to the above formula

So,

$= 438 \times \frac{1.333}{1.501}$

= 388.97 nm

We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come

5 0

3 years ago

When the dry-bulb reading of a thermometer is 20°C and the wet-bulb reading is 11°C, the relative humidity is approximately

8090 [49]

Answer:

You need to use charts or a calculator to get 56%

Explanation:

3 0

3 years ago

Solve the equation. 5x-5=10x+2

Veronika [31]

Answer:

x = - 1.4

Explanation:

-5=10x+2-5x (subtract 5x from both sides)

-5=5x+2 (simplify)

-5-2=5x (subtract 2 from both sides)

-7=5x (simplify)

x=-7/5 (divide both sides by 5)

x=-1.4 (simplify)

i would really appreciate getting a brainliest. anyways i hope this helped and have a great rest of your day/night!! :)

8 0

4 years ago

Read 2 more answers

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

or

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

or

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

or

I = $0.0109[1-2.717^{-1}]$

or

I = 0.00688 A

or

I = 6.88 mA

5 0

3 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago