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2 years ago

A kickball is kicked straight up at a speed of 22.4m/s. how high does it go

Physics

1 answer:

Kaylis [27]2 years ago

3 0

Answer:

the kickball goes straight up 25.6 meters

Explanation:

We are in the case of motion under constant (uniform) acceleration. This is an object with initial velocity pointing strictly upwards, under th acceleration of gravity which slows down its motion. The projectile keeps losing velocity as it moves upward until its velocity is zero when it stops moving upwards, and starts falling also under accelerated motion and gaining downward velocity as time goes by.

We use he formula for displacement under accelerated motion (due to gravity "g" opposing the initial velocity):

$y-y_0=v_0\,t-\frac{1}{2} g\,t^2$

and also the formula for the velocity in terms of time:

$v(t)=v_0-g\,t$

Using that the initial velocity is 22.4 m/s, g = 9.8 m/s^2 , and the fact that we want the time "t" at which the velocity v(t) = 0 (zero), we solve for "t" in the second equation, and use it to substitute for "t" in the first one:

$0=v_0-g\,t\\t=\frac{v_0}{g} \\\\\\y-y_0=v_0\,t-\frac{1}{2} g\,t^2\\y-y_0=v_0\,(\frac{v_0}{g})-\frac{1}{2} \,g\,(\frac{v_0}{g})^2\\y-y_0=\frac{v_0^2}{g} -\frac{1}{2}\,\frac{v_0^2}{g}\\y-y_0=\frac{1}{2}\,\frac{v_0^2}{g}\\y-y_0=\frac{1}{2}\,\frac{22.4^2}{9.8}\,\,m\\\\y-y_0=25.6\,\,m$

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A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d

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Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7 , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional magnetic flux is induced out of the page.

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2 years ago

a submarine moving directly upward in the water at constant speed. The weight of the summer and it is 500,000 N. The submarines

Alex787 [66]

The answe would be A

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2 years ago

You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a

gtnhenbr [62]

Answer:

The necessary separation between the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

$V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}$

Where;

d is the separation or distance between the two parallel plates;

$d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm$

Therefore, the necessary separation between the two parallel plates is 0.104 mm

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2 years ago

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in2 and weight of 100 lbf. The atmo

kompoz [17]

Answer:

ΔU=0.8834 Btu

Explanation:

Given data

Area of piston A=40 in²

The weight W=100 lbf

Atmospheric Pressure P=14.7 lbf/in²

Work added E=3 Btu

The change in elevation Δh=1 ft =12 inch

To find

Change in internal energy of the gas ΔU

Solution

For Piston

ΔPE=| W+(P×A)×Δh |

ΔPE=| 100+(14.7×40)×12 |

ΔPE=8256 lbf.in

ΔPE=8256×0.000107

ΔPE=0.8834 Btu

From law of conservation of energy then ,the charging in the potential energy of the piston is made by exerting force by gas

Wgas= -ΔPE

Wgas= -0.8834 Btu

For the gas as a system and by applying first law of thermodynamics

Q-W=ΔU

0-(-0.8834 Btu)=ΔU

ΔU=0.8834 Btu

6 0

2 years ago

In a later chapter we will be able to show, under certain assumptions, that the velocity v(t) of a falling raindrop at time t is

lianna [129]

Answer:

$\lim_{t \to \infty} v(t) =vT$

Explanation:

Using distributive propierty:

$v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}$

So:

$\lim_{t\to \infty} vT-e^{-gt}$

The limit of the sum of two functions is equal to the sum of their limits, therefore:

$\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}$

The limit of a constant function is the constant, hence:

$\lim_{t\to \infty} vT=vT$

Now, let's solve the other limit:

$\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}$

The limit of a constant times a function is equal to the product of the constant and the limit of the function, so:

$\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)$

$-g(\infty)=-\infty$

Therefore:

$e^{(-\infty)} =0$

Finally:

$\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT$

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2 years ago