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deff fn [24]
3 years ago
14

Chose terms and put in answer spaces Answer box- difference direction displacement distance final initial reference point Senten

ce: an objects position is its__ and __ from a__. Sentence 2:__ is__ between its __ and __ positions
Physics
1 answer:
katrin [286]3 years ago
5 0

Answer:

1. An objects position is its <u>distance</u> and <u>direction</u> from a <u>reference point</u>.

2.<u>Displacement</u> is <u>difference</u> between its <u>final</u> and <u>initial</u> positions.

Explanation:

For any position we required a reference point and the distance between the reference point and the object.

For example:

an object is 40 m away from a table.

Here the <u>distance is 40 m</u> from a reference point that is <u>a table.</u>

Displacement is the difference between its final and initial position.

For example

A person walk 20 km and then returned back 10 km.

Here distance traveled by a person is 30 km. But displacement is (20 - 10) that is the difference between final and initial position.

So displacement is 10 km.

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A long pipe of outer radius R 1 = 3.70 cm and inner radius R 2 = 3.15 cm carries a uniform charge density of 1.22 mC/m 3 . Assum
7nadin3 [17]

Answer:

E = 4.72 * 10⁻⁶ Nm²

Explanation:

Parameters given:

Outer radius, R = 3.70cm = 0.037m

Inner radius, r = 3.15cm = 0.0315m

Permittivity of free space, ε₀ = 8.85 * 10⁻¹² C/Nm²

Charge density: 1.22 * 10⁻³ C/m³

The question requires that we solve using Gauss law which states that the net electric field through a closed surface is proportional to the enclosed electric charge.

Hence,

E = Q/Aε₀

Charge Q is given as

Q = ρπ(R² ⁻ r²)L

A = 2π(R - r)L

E = [ρπ(R² ⁻ r²)L]/[2π(R - r)ε₀L]

Using difference of two squares,

(R² ⁻ r²) = (R + r)(R - r)

E =[ρ(R + r)]/(2ε₀)

E = [1.22 * 10⁻³ *(0.0370 + 0.0315)]/(2 * 8.85 * 10⁻¹²)

E = 4.72 * 10⁻⁶ Nm²

6 0
4 years ago
Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by
mrs_skeptik [129]
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R  as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick T_R=T whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that T_L=-T

8 0
3 years ago
What equation p=m x v used to determine
OverLord2011 [107]

Answer:

The Momentum Calculator uses the formula p=mv, or momentum (p) is equal to mass (m) times velocity (v). The calculator can use any two of the values to calculate the third.

Explanation:

5 0
3 years ago
Observe: Air pressure is equal to the weight of a column of air on a particular location. Air pressure is measured in hectopasca
babymother [125]

Answer:

A. When moving towards a high pressure center the pressure values ​​increase in the equipment

B. This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

4 0
3 years ago
Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge
Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

5 0
4 years ago
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