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3 years ago

12

The electromagnetic waves that have the lowest frequencies are called

1 answer:

o-na [289]3 years ago

4 0

A. Radio waves
Have the lowest frequencies

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A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold

salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did

dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

Initial Velocity (Vo) = 7 m/s
Acceleration (a) = 3 m/s²
Final Velocity (Vf) = 22 m/s
Time (t) = ?

Use formula:

$\boxed{t=\frac{Vf - Vo}{a}}$

Replace:

$\boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}$

Solve the subtraction of the numerator:

$\boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}$

It divides:

$\boxed{t=5\ s}$

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

8 0

2 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

2 years ago

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$

6 0

2 years ago

You see a total eclipse of the sun. Three Saros cycles later, you find yourself at the same location.

alukav5142 [94]

Answer:

The correct answer is B. You would not have to travel far to see another total eclipse .

Explanation:

it is said that the way to predict or know beforehand if there is a lunar or solar eclipse is just to know the what is called the saros which are periods of exactly 223 synodic months and the eclipses are repeated or occurred in the same location every 3 Saros cycles.

5 0

2 years ago