Determine one way you can contribute to water in the atmosphere in your day-to-day activities pleaseeeee helppp

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

PtichkaEL [24]

This is True

Kinetic energy is the energy of motion. The bicyclist is in motion as he pedals up the tall hill. Therefore, the bicyclist contains kinetic energy.

4 0

3 years ago

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When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca

ratelena [41]

Positively charged protons in the nucleus of the gold atom .... rutherford scattering ???

3 0

3 years ago

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An electric motor has an effective resistance of 22 22 and an inductive reactance of 72 12 when working under load. The voltage

Alecsey [184]

Answer:

Current amplitude, I = 5.57 A

Explanation:

It is given that,

Effective resistance, R = 22 ohms

Inductive reactance, L = 72 ohms

Voltage of the alternating source, V = 420 volt

The total impedance of the RL circuit is given by :

$Z=\sqrt{R^2+L^2}$

$Z=\sqrt{(22)^2+(72)^2}$

Z = 75.28 ohms

From Ohm's law,

$V=I\times Z$

$I=\dfrac{V}{Z}$

$I=\dfrac{420}{75.28}$

I = 5.57 A

So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.

6 0

3 years ago

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

3 years ago

A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At

Bad White [126]

Answer:

Electric field is zero at point 4.73 m

Explanation:

Given:

Charge place = 50 cm = 0.50 m

change q1 = 5 µC

change q2 = 4 µC

Computation:

electric field zero calculated by:

$E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\$

Where electric field is zero,

First distance = x

Second distance = (x-0.50)

So,

E1 = E2

$k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\$

$\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\$

x = 0.263 or x = 4.73

So,

Electric field is zero at point 4.73 m

8 0

3 years ago