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3 years ago

What type of force is Ft?

Physics

2 answers:

Katena32 [7]3 years ago

8 0

Answer:

Force due to tension

Explanation:

Usually Ft, or F $_{t}$ , denotes force due to tension or tension force. An example of tension force would be if you were pulling an object on a string - the tension force is present within the string and causes the string not to break.

ipn [44]3 years ago

4 0

Answer:

=0.1382550kgf.

Explanation:

gravitational metric system

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4.333333 kilometers an hour

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nydimaria [60]

Complete Question

The diagram of with this question is shown on the first uploaded image

Answer:

The value is $v = -6.543 \^ i + 9.47 \^ j + 0 \^ k$

Explanation:

From the question we are told that

The mass of the rock is $m = 0.250 \ kg$

The length of the string is $L = 0.75 \ m$

The angle the string makes horizontal is $\theta = 11.9^o$

The angle which the projection of the string onto the xy -plane makes with the positive x-axis is $\phi = 34.6^o$

The angular velocity of the rock is $w = 2.50 rev/s = 2.50 * 2\pi =15.7 \ rad/s$

Generally the radius of the circle made by the length of the string is mathematically represented as

$r = L cos(\theta )$

=> $r = 0.75 cos(11.9 )$

=> $r = 0.734 \ m$

Generally the resultant tangential velocity is mathematically represented as

$v__{R}} = w * r$

=> $v__{R}} = 15.7 *0.734$

=> $v__{R}} = 11.5 \ m/s$

Generally the tangential velocity along the x-axis is

$v_x = -v__{R}} * sin(\phi)$

=> $v_x =- 11.5 * sin(34.6)$

=> $v_x = -6.543 \ m/s$

The negative sign show that the velocity is directed toward the negative x-axis

Generally the tangential velocity along the y-axis is

$v_y = v__{R}} * cos(\phi)$

=> $v_y = 11.5 * cos(34.6)$

=> $v_y = 9.47 \ m/s$

Generally the tangential velocity along the y-axis is

$v_z = v__{R}} * cos(90)$

=> $v_z = 0 \ m/s$

Generally the tangential velocity at that instant is mathematically represented as

$v = -6.543 \^ i + 9.47 \^ j + 0 \^ k$

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3 years ago

An infinite plane of charge has a surface charge density of 5 µC/m2 . How far apart are the equipotential surfaces whose potenti

Lostsunrise [7]

Answer:

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Explanation:

We have given charge surface charge density $\rho _s=5\mu c/m^2=5\times 10^{-6}\mu c/m^2$

We know that electric field due to surface charge density is given by

$E=\frac{\rho _S}{2\epsilon _0}=\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}=2.824\times 10^5Volt/m$

We have given potential difference V = 105 volt

We know that potential difference is given by $V=Ed$

So $105=2.824\times 10^5\times d$

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