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Triss [41]
3 years ago
12

When the sun and moon pull at right angles to the earth wat kinda tide can yu expect

Physics
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

A neap tide. Hope this helps

Explanation:

You might be interested in
1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and
saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

   Centripetal acceleration  = \frac{v^{2} }{r}

   v is the velocity of the body

    r is the radius of the circle

  putting in the parameters:

   Centripetal acceleration = \frac{4^{2} }{0.75}

    Centripetal acceleration = 10.67m/s²

Centripetal force = m  \frac{v^{2} }{r}

          m is the mass

 Centripetal force = mass x centripetal acceleration

                              = 3 x 10.67

                              = 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

4 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
8090 [49]

Answer:

\Delta L = 4.2 m

Explanation:

As per the formula of thermal expansion we know that

L = L_o(1 + \alpha\Delta T)

so here we will have

L_o = 1400 m

\alpha = 12 \times 10^{-6} per ^oC

\Delta T = 250 degree C

so here change in the length of the rod is given as

\Delta L = L - L_o

\Delta L = L_o \alpha \Delta T

\Delta L = 1400 (12 \times 10^{-6})(250)

\Delta L = 4.2 m

8 0
3 years ago
Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How
zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

angle = \frac{1 Parsec}{distance}

so we have

distance = \frac{1 Parsec}{angle}

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

d_a = \frac{1 Parsec}{1} = 1 Parsec

distance of star B is given as

d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec

So star A is closer than star B

7 0
3 years ago
Which circuit is a series circuit ?? plz help
gregori [183]

In a series circuit, all of the components are connected in the same 'loop' and the current only has one direction/path it can flow through.

In the first three options, the current has multiple paths it can go through. So these three circuits are parallel and not series.

In the last option, the current only has one path where it can flow through, so that circuit is in series.

So Circuit <u>D </u>is a series circuit.

----------------------------------------

Answer

Circuit D

7 0
3 years ago
Read 2 more answers
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