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2 years ago

Plz help will give 26 points !and be mark as brainliest !

Physics

2 answers:

Scilla [17]2 years ago

6 0

The answer for question #3 is Relief!!!!

Dimas [21]2 years ago

4 0

Answer:

1. The "contour interval' identifies the standard difference in elevation on a topography graph. It shows the difference in elevation of highest and lowest points on a topography map. In topographic map every fifth contour line is drawn, wider than the contour lines, called as contour index.

2. Geologist usually study all the materials that make up the earth. He can focus on different rock samples from earth's surface. the geologist works to understand history of the planet where we live on, to predict the future by explaining current occurrences.

3. The difference between the top of the plateau and the lowest point in the area of 250 m measurement is called elevation. Elevation is the highest point in the area.

4. Damage from earthquakes is determined by "Ritcher scale". Seismic waves are the vibrations occurs from earthquakes that can travel through the earth. These vibrations can be detected by the Ritcher scale.

You might be interested in

Sort the sentences based on whether they describe radio waves, visible light waves, or both.

valentina_108 [34]

The electromagnetic spectrum includes a continuous spectrum of wavelengths that include:

Radio waves, microwaves, infrared light, visible, ultraviolet, X-rays, gamma rays

The wavelength decreases from radio waves to gamma rays, whereas the energy increases along the same direction.

In the given example, radio waves have a lower energy and higher wavelength than visible light. The latter can be perceived by the human eye, whereas radio waves are not visible to the human eye.

1) They have colors = visible light

2) They can travel in a vacuum = both

3) They have energy = both

4) They’re used to learn about dust and gas clouds = radio waves

5) They’re used to find the temperature of stars = visible light

6)They’re invisible = radio waves

6 0

3 years ago

What type of energy does an object gain as it is lifted at a constant speed?

marshall27 [118]

D. Gravitational potential

7 0

3 years ago

A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement

Igoryamba

Answer:

Angular displacement=68.25 rad

Explanation:

Circular Motion

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

$\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}$

The angular displacement is given by:

$\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}$

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

$\displaystyle \alpha=\frac{6-13.5}{7}$

$\displaystyle \alpha=\frac{-7.5}{7}$

$\displaystyle \alpha=-1.071 \ rad/s^2$

Thus, the angular displacement is:

$\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}$

$\displaystyle \theta=94.5-26.25$

$\boxed{\displaystyle \theta=68.25\ rad}$

Angular displacement=68.25 rad

3 0

3 years ago

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080

deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

$I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}$

now we have

$M = 0.120 kg$

$m_1 = 0.02 kg$

$m_3 = 0.08 kg$

now we have

$I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}$

so we have

$I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162$

$I = 0.02835$

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

$\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}$

$\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)$

$\omega = 4.32 rad/s$

Now speed of 0.08 kg mass when it reaches to bottom point is given as

$v = \omega \frac{L}{2}$

$v = 4.32 (0.45)$

$v = 1.94 m/s$

3 0

3 years ago

What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = 7.6 μc, twice the charge o

Anna [14]

Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.

dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m

I answeared your question can you answear my question pleas

6 0

3 years ago