If you know the distance of an earthquake epicenter from three seismic stations, how can you find the exact location of the epicenter of the earthquake.

Answers

Answer 1

Answer:

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

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The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!

Answers

Answer:

$F_N=1234.8N$

Explanation:

Hello.

In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:

$F_N=63kg*9.8m/s^2+500N\n\nF_N=617.4N+500N\n\nF_N=1234.8N$

Best regards.

A large, massive object collides with a stationary, smaller object on an ice rink. If the large object transfers all of its momentum to the smaller object, which statement below describes the velocity of the smaller object after the collision? A. It will move faster than the large object was moving initially.

B. It won't move.

C. It will move at the same speed that the large object was moving initially.

D. It will move slower than the large object was moving initially.

Answers

Answer:

a ut will move faster than the large object was moving initially

Answer: It will move faster than the large object was moving initially.

Explanation:

Three point mass particles are located in a plane: a. 3.77 kg located at the origin
b. 6.7106 kg at [(5.72 cm),(11.44 cm)]
c. 2.46181 kg at [(16.7024 cm),(0 cm)].

How far is the center of mass of the three particles from the origin? Answer in units of cm

Answers

The distance of the center of mass of the three particles from the origin is 6.1428 cm and 5.9316 cm.

Calculation of the distance:

Since

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Now here we assume x and y be the coordinates with respect to the centre of mass.

So,

We know that

$x = (m_1x_1+m_2x_2+m_3x_3)/(m_1+m_2+m_3)\n\n = (3.77* 0 + 6.7106 * 5.72 + 2.46181 * 16.7024)/(3.77 + 6.7106 + 2.46181)$

= 6.1428 cm

Now

$y = (m_1y_1+m_2y_2+m_3y_3)/(m_1+m_2+m_3)\n\n = (3.77* 0 + 6.7106 * 11.44 + 2.46181 * 0)/(3.77 + 6.7106 + 2.46181)$

= 5.9316 cm

Learn more about mass here: brainly.com/question/16876455

Answer:

Explanation:

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Let x and y be the coordinates of centre of mass.

$x = (m_(1)x_(1)+ m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))$

$x = (3.77* 0+ 6.7106* 5.72 + 2.46181* 16.7024)/(3.77+6.7106+2.46181)$

x = 6.1428 cm

$y = (m_(1)y_(1)+ m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))$

$y= (3.77* 0+ 6.7106* 11.44 + 2.46181* 0)/(3.77+6.7106+2.46181)$

y = 5.9316 cm

Which of the following is a good example of a contact force?ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up

Answers

Answer:

A bridge suspended by cables

Explanation:

Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.

A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]