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evablogger [386]
3 years ago
6

Please talk about the earths layers and plate tectonics when answering

Physics
1 answer:
Alinara [238K]3 years ago
7 0

Answer:

zscDSGVDTBHYHJMUK,K.

Explanation:

zxDVGFVHJ,K.L/;KJHGFDSAQdwgyhu'[;

['piou6t54321345367687890-568+

635836.

69

32658

":lkjhgfdsaqDDGTFHUHKL;'P

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How does a frequency of a wave relate to the energy of a wave
ch4aika [34]
The higher the frequency of a wave the more energy the wave has.
4 0
3 years ago
Read 2 more answers
A car is traveling at some speed when it accelerates at 6 m/s2 for 3 seconds. If the car travels 39 meters in this time, how fas
Alchen [17]

Answer:

4 m/s

Explanation:

Given:

Δx = 39 m

a = 6 m/s²

t = 3 s

Find: v₀

Δx = v₀ t + ½ at²

39 m = v₀ (3 s) + ½ (6 m/s²) (3 s)²

v₀ = 4 m/s

6 0
3 years ago
point) A tank in the shape of an inverted right circular cone has height 5 meters and radius 4 meters. It is filled with 4 meter
zysi [14]

Answer:

W = 907963.50 J = 907.96 J

Explanation:

Note: Refer to the figure attached

Now, from the figure we have similar triangles ΔAOB and ΔCOD

we have

\frac{5}{4}=\frac{x}{r}

or

r=\frac{4x}{5}

Now, the work done to empty the tank can be given as:

W = \int\limits^4_0 {(5-x)\rho\times g A} \, dx

or

W = \int\limits^4_0 {(5-x)1080\times 9.8 (\pi r^2)} \, dx

or

W = \int\limits^4_0 {(5-x)\times10584\times (\pi (\frac{4x}{5})^2)} \, dx

or

W = 6773.76\pi\int\limits^4_0 {(5-x)x^2)} \, dx

or

W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0

or

W = 6773.76\pi[\frac{128}{3}]

or

W = 907963.50 J = 907.96 J

7 0
3 years ago
Bats can detect small objects such as insects that are of a size on the order of a wavelength. if bats emit a chirp at a frequen
crimeas [40]
The smallest size of the insect that the bats can detect corresponds to the wavelength of the chirp they emit.

Their chirp has a frequency of
f=69.3 kHz=69.3 \cdot 10^3 Hz
and the speed of the chirp is equal to the speed of sound in air:
v=330 m/s
Therefore the wavelength of the chirp is
\lambda= \frac{v}{f}= \frac{330 m/s}{69.3 \cdot 10^3 Hz}=4.76 \cdot 10^{-3} m
which corresponds to a size of 4.76 mm.
6 0
4 years ago
"a 10 kg rock is pushed off the edge of a bridge 50 meters above the ground. what was the kinetic energy of the rock at the midw
valentinak56 [21]
Let's call h the initial height of the rock (h=50 m, the height of the bridge).

Initially, the rock has only gravitational potential energy, which is given by
U_i=mgh
where m=10 kg is the mass of the rock while g is the gravitational acceleration. So, the total mechanical energy of the rock at this point is
E_i = U_i = mgh=(10 kg)(9.81 m/s^2)(50 m)=4905 J

At midway point of its fall, its height is \frac{h}{2}, so its potential energy is
U_f = mg \frac{h}{2} = (10 kg)(9.81 m/s^2)(25 m)=2452.5 J
But now the rock is also moving by speed v, so it also has kinetic energy:
K_f =  \frac{1}{2}mv^2
So the total energy at the midway point of the fall is
E_f = U_f + K_f (1)

The mechanical energy must be conserved, so E_i = E_f, so we can rewrite (1) and solve it to find the kinetic energy of the rock at midway point of its fall:
E_i = U_f + K_f
K_f = E_i - U_f = 4905 J - 2452.5 J=2452.5 J
8 0
3 years ago
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