A wave traveling in water has a frequency of 250 Hz and a wavelength of 6.0 m. What is the speed of the wave

Many kinds of analysis in the physical sciences are helped by identifying quantities that are constant. It is usual to begin by

AysviL [449]

Answer:

As, there are different physical quantities which are required to be measured in a more proper way to define the circuits properties.

As, there is current,I which is the more directed flow of electrons inside the circuit.
Along, with which the voltage,V or in other words the potential difference inside the circuit is also taken under consideration.
And most importantly the resistance's,R value inside the circuit defines the parameters inside the circuit.
All of these values are then determined or analyzed with the help of the digital multi-meter or simply the DMM.

8 0

3 years ago

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts

wolverine [178]

Answer:

0.304 m/s2

Explanation:

If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.

We can use Newton's 2nd law to calculate the net acceleration of the system

$a = F/m = 7 / 23 = 0.304 m/s^2$

5 0

3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Temperature is a measure of work.<br> True<br> False

Veseljchak [2.6K]

The answer is false.

5 0

3 years ago

The graph below shows the conservation of energy for a skydiver jumping out

frez [133]

Answer:

The correct option is:

B) Kinetic Energy

Explanation:

We know that if a body is placed at a certain height, it possesses Potential Energy, which is represented by 'mgh'. In this case, when the skydiver is present in the plane, before jumping, he has potential energy as he is at height 'h'.

As Kinetic energy is given as '(1/2)mv²' dependent on velocity of the object, when the skydiver jumps of the plane, his height starts decreasing, which decreases his Potential Energy. As energy can neither be created or destroyed, but is converted to one form or another, all this Potential energy starts to convert into Kinetic energy. As Potential Energy decreases with distance, Kinetic energy increases. Hence, line B represents Kinetic Energy.

3 0

3 years ago