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garik1379 [7]
3 years ago
5

Three point charges are fixed in place in a right triangle, as shown in the figure.

Physics
1 answer:
8090 [49]3 years ago
4 0
Oh gosh oh I see it in my life face and
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Transverse waves travel with a speed of 20 m/s on a string under a tension 0f 6.00 N. What tension is required for a wave speed
Aleksandr [31]

Answer:

T_2=13.5\ N

Explanation:

Given that,

Speed of transverse wave, v₁ = 20 m/s

Tension in the string, T₁ = 6 N

Let T₂ is the tension required for a wave speed of 30 m/s on the same string. The speed of a transverse wave in a string is given by :

v=\sqrt{\dfrac{T}{\mu}}........(1)

T is the tension in the string

\mu is mass per unit length

It is clear from equation (1) that :

v\propto\sqrt{T}

\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}}

T_2=T_1\times (\dfrac{v_2}{v_1})^2

T_2=6\times (\dfrac{30}{20})^2

T_2=13.5\ N

So, the tension of 13.5 N is required for a wave speed of 30 m/s. Hence, this is the required solution.

5 0
3 years ago
Why do we experience a high tide twice a day?
Veseljchak [2.6K]

Coastal areas experience two low tides and two high tides every lunar day, or 24 hours and 50 minutes. The two tidal bulges caused by inertia and gravity will rotate around the Earth as the moons position changes. These bulges represent high tides while the flat sides indicate low tides.

Over the course of 1 day, the position of the moon does not change very much compared to the rotation of the earth. As the earth rotates below the moon, one point on the earth will go through all levels of tide as the day passes by. Strongly attracted, middling, weakly attracted, and then middleagain. From our perspective it looks like "high, low, high, low." Or equivalently you can think about how the points below the moon and opposite the moon will be high tide, and as the earth rotates, those areas will change.

7 0
3 years ago
Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a fo
Dafna1 [17]

Answer: 75 N to the right

Explanation:

5 0
3 years ago
The velocity of an object is equal to the distance divided by time. The equation is velocity = distance/time. If you wanted to c
Korolek [52]

Answer: C) divide: distance ÷ velocity

Explanation:

The velocity V equation is distance d divided by time t:

V=\frac{d}{t}

If we isolate t we will have:

t=\frac{d}{V}

Hence, the correct option is C: distance divided by velocity.

7 0
3 years ago
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
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