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In what ways can students use time management? Check all that apply.

Korvikt [17]

Answer: the answer is A, C, E

Explanation:

Time management is apart of your study schedule so you will need to determine when to study, same goes with how to use ur study time, lastly it helps create ur study schedule to help find out what time you are free to study. :)

7 0

3 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

How large is the area of water found on Mars?

ehidna [41]

More than five million cubic kilometers of ice have been identified.

8 0

3 years ago

Read 2 more answers

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

gizmo_the_mogwai [7]

Answer:

Explanation:

First of all we shall find the velocity at equilibrium point of mass 1.2 kg .

It will be ω A , where ω is angular frequency and A is amplitude .

ω = √ ( k / m )

= √ (170 / 1.2 )

= 11.90 rad /s

amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

= .5355 m /s

At middle point , no force acts so we can apply law of conservation of momentum

m₁ v₁ = ( m₁ + m₂ ) v

1.2 x .5355 = ( 1.2 + .48 ) x v

v = .3825 m /s

= 38.25 cm /s

Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

A₁ = .0379 m

New amplitude is .0379 m

7 0

3 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

3 years ago