Which type of biological molecule would contain fats?

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

Mazyrski [523]

I thinks its He uses proof to show the evidence is relevant. But im not totally positive on it hope this helps

8 0

2 years ago

Read 2 more answers

You use 42 J of energy to move a 6.0 N object in 2 seconds. How far did you move it?

bogdanovich [222]

22 that's how much I moved

7 0

2 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

2 years ago

A balloon is ascending at 12.4m/s at a height of 81.3m above the ground when a package is dropped. a) How long did it take to re

tankabanditka [31]

Answer:

3secs

Explanation:

Given the following parameters

height H= 81.3m

Velocity v = 12.4m/s

Required

Time it take to reach the ground

Using the equation of motion

H = ut+1/2gt²

81.3 = 12.4t + 1/2(9.8)t²

81.3 = 12.4t + 4.9t²

4.9t² + 12.4t - 81.3 = 0

Using the general formula to find t

t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)

t = -12.4±√153.76+1593.48/2(4.9)

t = -12.4±√1747.24/9.8

t = -12.4+41.8/9.8

t = 29.4/9.8

t = 3secs

Hence it took 3secs to reach the ground

5 0

2 years ago

A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.

hichkok12 [17]

Answer:

a) $\Delta U_g=-5.3kJ$

b) $K=0.27kJ$

c) $F_f=0.45kN$

Explanation:

the gravitational potential energy is given by:

$U_g=m.g.h\\$

$\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ$

The kinetic energy is given by:

$K=\frac{1}{2}m.v^2\\$

the initial kinetic energy is zero because the motion started from rest, so:

$K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ$

applying the conservation of energy theorem:

$U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ$

The work done by the friction force is given by:

$W_f=F_f.h.cos(\theta)\\$

the angle of the force is 180 degrees because it's against the movement:

$F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN$

8 0

3 years ago