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Schach [20]
3 years ago
12

The period of a wave is the reciprocal of its ______________.

Physics
1 answer:
Hunter-Best [27]3 years ago
6 0
Frequency

hope it helps !  
You might be interested in
Which of the following is an example of a projective test?
IceJOKER [234]

Some examples of projective tests are the Rorschach Inkblot Test, the Thematic Apperception Test (TAT), the Contemporized-Themes Concerning Blacks test, the TEMAS (Tell-Me-A-Story), and the Rotter Incomplete Sentence Blank (RISB).

Some examples of projective tests are the Rorschach Inkblot Test, the Thematic Apperception Test (TAT), the Contemporized-Themes Concerning Blacks test, the TEMAS (Tell-Me-A-Story), and the Rotter Incomplete Sentence Blank (RISB).

6 0
3 years ago
Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC
Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

q_1=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 1)

q_2=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N

5 0
3 years ago
A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-
Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0
3 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu
tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

where

\Delta V is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

E=5.00\cdot 10^6 V/m

d = 2.50 mm = 2.50\cdot 10^{-3} m

Substituting, we find

\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V

3 0
3 years ago
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