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GuDViN [60]
3 years ago
10

What type of appeal does Gore mostly use to persuade the audience?

Physics
2 answers:
Mazyrski [523]3 years ago
8 0
I thinks its He uses proof to show the evidence is relevant. But im not totally positive on it hope this helps
BARSIC [14]3 years ago
3 0

Answer:

He always try to prove that the evidence is relevant.

Explanation:

The Gore speeches are mostly emotional speeches he speaks mostly about those topics which are very heart touching or influencing means which are emotional and turn the public emotionally in his favor.

And he also uses that kind of evidences which are very relevant to the situation and always try to prove his thoughts right and persuade the audience.

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A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times
Ira Lisetskai [31]

Answer:

9R

Explanation:

We know that the resistance is R=\rho *\frac{L}{A}.

If we stretch the wire to a new length L2 = 3L, the cross-sectional area will also change. If the cross-sectional area doesn't change throughout the wire, we can say that:

Volume = L*A = 3L * A2    being A2 the new area after stretching the wire.

Since the volume remains the same we conclude that A2 = A/3

With this information, we calculate the new resistance:

R2=\rho *\frac{L2}{A2}=\rho *\frac{3*L}{A/3}=\rho * 9 * \frac{L}{A}

Since R=\rho *\frac{L}{A}, and by simple inspection of the previous equation, we get:

<em>R2 = 9*R</em>

6 0
3 years ago
Three polarizing filters are shown in the figure below. The alignment of each filterâs transmission axis is shown, as measured f
lidiya [134]

Answer: B.

Explanation:

Hope help :p

5 0
2 years ago
A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli
fredd [130]

Answer:

Part a)

A = 1.78 m

Part b)

f = 2 rev/s

Explanation:

Part A)

As we know that time period of the motion is given as

T = 2.68 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.68}

\omega = 2.34 rad/s

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

mg = m\omega^2 A

so we have

A = \frac{g}{\omega^2}

A = \frac{9.81}{2.34^2}

A = 1.78 m

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

mg = m\omega^2 A

g = \omega^2 (0.0623)

\omega = 12.5 rad/s

so now we have

2\pi f = 12.5

f = 2 rev/s

3 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
How does the mass affect the speed as it rolls down a slope?
VikaD [51]
As we use the Kinetic energy and the equation is 1/2mv^2, changing its mass will change its speed and its energy. So more mass, more speed more energy. also the gravitational potential energy; mass x gravity x height; more mass and more height more speed as it go down to the slope! Hope it helps!
6 0
3 years ago
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