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3 years ago

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

1 answer:

8 0

From our physics class, we know that Work is a product of Force and Displacement. The force must be the force in the same direction with the displacement. So force and displacement must be parallel.

A. How much work in J does the string do on the boy if the boy stands still?

Since we know that: Work = Force * Displacement

and in this case displacement is equal to zero, therefore Work must also be equal to zero.

Work = 0

B. How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite?

Now are given a displacement of 11 m which is a horizontal distance. Therefore we must first calculate the horizontal component of force:

Fx = 4.5 N * cos 30

Fx = 3.897 N

So work is:

Work = 3.897 N * 11 m

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Give an example of a physical change and the property associated with that change. Give an example of a chemical change and the

dusya [7]

Answer:

A physical change is an ice cube melting into water. The property is melting point. Another is a banana changing from yellow to brown and the property is color. A chemical change example is a pot or a tool rusting and turning from silver to a reddish-brown. It is also a color change, but it is different because the color changed chemically and not just by looks.

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3 years ago

In which reaction are the atoms of elements rearranged?

NNADVOKAT [17]

Answer:

D. Hydrogen combines with oxygen to form water.

Explanation:

Hydrogen combining with oxygen to form water is a typical example of chemical reaction.

During a chemical reaction, atoms of elements are rearranged. Most chemical reactions obey the law of conservation of mass which states that "matter is neither created nor destroyed in a chemical reaction but atoms are simply rearranged".

The other choices given are nuclear reactions. In such reactions, atoms are not rearranged but are simply destroyed and made in the process.

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3 years ago

A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If

sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)= $50 \Omega$

Voltage(V)=15 V

Time constant( $\tau$ )= $\frac{L}{R}$

$\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}$

current $i_0=\frac{V}{R}$

$i_0=\frac{15}{50}=0.3 A$

Current as a function of time is given by

$i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )$

$i=0.3\times 0.9998$

i= 299.95 mA

6 0

4 years ago

g The ball has a mass of 150 grams. If the ball has a positive charge, what initial velocity must it have at the 200-V level for

Aleonysh [2.5K]

Answer:

u_o = 12.91 m/s

Explanation:

Given:

- mass of the ball m = 0.15 kg

- Spacing between each equipotential line s = 1.0 m

- Charge on the ball is q = 10 mC

Find:

- What initial velocity must it have at the 200-V level for it to reach its maximum height at the 700-V level

Solution:

- Notice that the Electric field lines are directed from Higher potential to lower potential i.e V = + 700 level to V = 200 level. The gravitational acceleration also acts downwards.

- We will compute a resultant acceleration due to gravity and Electric Field as follows:

g' = g + F_e / m

- Where F_e is the Electrostatic Force given by:

F_e = E*q

- Electric field strength is given by:

E = V / d

- Hence,

g' = g + (dV*q/ d*m)

- Input values:

g' = 10 + (500*10^-2 / 5*0.15)

g' = 16.6667 m/s^2

- Now use one of the equation of motions in y-direction:

h_max = u_o ^2 / 2*g'

- Input values where h_max = 5 m

5*2*g' = u_o ^2

u_o = sqrt (166.667)

u_o = 12.91 m/s

4 0

4 years ago

harold threw a baseball 3 meters harold is 2 meters tall. how fast did he throw the ball and how long was it in the air

frozen [14]

Answer:

u = 7.74 m/s and t = 1.82 s

Explanation:

Given that,

Harold threw a baseball 3 meters.

Harold is 2 meters tall.

The vertical distance = 3m

The final velocity at the maximum height is 0.

Let u is the initial velocity of the baseball. Using the first equation of motion to find it as follows :

$v^2-u^2=2as$

Here, a = -g

$v^2-u^2=-2gs\\\\u=\sqrt{2gs} \\\\=\sqrt{2\times 10\times 3} \\\\=7.74\ m/s$

He has thrown it with the speed of 7.74 m/s

Let t is the time taken by the ball to fall on the ground.

Harold is 2 meters tall. Using second equation of motion to find it.

u = -7.74 (as downward direction is taken as positive)

$s=ut+\dfrac{1}{2}at^2\\\\2=-7.74t+5t^2\\\\t=1.82\ s$

Hence, this is the required solution.

5 0

3 years ago