All categories

3 years ago

Compare and contrast an earthquake and a tsunami.

Physics

1 answer:

Scorpion4ik [409]3 years ago

4 0

Answer:

An earthquake is a trembling movement of the earth's crust. These tremors are generally caused by shifts of the plates that make up the earth's surface. ... A tsunami (pronounced soo-NAHM-ee) is a series of huge waves that occur as the result of a violent underwater disturbance, such as an earthquake or volcanic eruption.

Explanation:

You might be interested in

Distance measurements based on the speed of light; used for objects in space

Vika [28.1K]

Answer : Yes, distance measurements based on the speed of light used for objects in space.

Explanation : A light year is measurement of distance that light travel in a one year.

In a one year light travels 9460000000000 kilometer.

We know that, speed of light is $3\times10^{8}\ m/s$

and time is 31536000 seconds in 1 year

so, distance = speed of light X time

Now, the light year is $9.4608\times10^{15} meter$

Example : The nearest star to earth is about 4.3 light year away.

4 0

3 years ago

A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a

sattari [20]

Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

11^2 + 11^2 = c^2

121 + 121 = c^2

242 = c^2

c = 15.56

Rounding the answer makes it 16 km for the hiker's magnitude of displacement.

5 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago

What accounts for most of the mass in the universe?

aev [14]

Around 80 percent of the mass of the universe is made up material known as "Dark matter". It does not emit light or energy but the influence of it can be detected or observed gravitationally. Motions of stars and galaxy tell us how much mater there is, but somehow the speed of rotation of galaxy does not add up to its mass alone, there is a certain amount of matter really not accounted for. Dark matter maybe made up of non-baryonic matter, or perhaps what scientist called the WIMPS or (weakly interacting massive particles.)

5 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago