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3 years ago

14

There are several events listed below some of the events are related to tectonic plates activity mark each that is associated wi

th plate activity (4 points)
A. Earthquakes
B. Hurricane
C. Tsunamis
D. Mountain building/ formation

1 answer:

iVinArrow [24]3 years ago

5 0

A because an earthquake is shaking of tectonic plates

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A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters

emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

$F = \frac{mv^{2} }{r}$

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = $2\pi r$

Work on a particle = 2πrF = $2\pi mv^{2}$

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

3 0

3 years ago

2. A 75 kg runner accelerates from 0.00 m/s to 10.0 m/s in 1.5 seconds.

raketka [301]

A :-) A.a.) Given - u = 0.00 m/s
v = 10.0 m/s
t = 1.5 sec
m = 75 kg
Solution -
a = v - u by t
a = 10 - 0 by 1.5
a = 10 by 1.5
a = 6.6 m/s^2

A.b.) sorry ! I don’t no how to do this question

3 0

3 years ago

A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of

jenyasd209 [6]

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

7 0

3 years ago

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the

Eduardwww [97]

Answer: 1.14 N

Explanation :

As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:

Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3 m3. 9.8 m/s2

Fb = 1.34 N

In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

Fg = (mb +mhe) g

The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg

Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

6 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago