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3 years ago

Infer whether a system can have kinetic energy and potential energy at the same time

1 answer:

7 0

A system can have kinetic and potential energy, for example, an object falling but it hasn't reached the ground. It's still in the air, and has "potential" to go faster or hit something on it's way down, but since it's moving it also has kinetic energy.

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Help me, Ive been stuck for 15 minutes

natita [175]

The answer is b) venus because the period rotation is also referred to as a day so if venus’ period of rotation is 243 [earth] days and its year is 225 [earth] days, then the period of rotation is longer.

5 0

3 years ago

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0

3 years ago

Please help on this one?

kupik [55]

Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V

7 0

3 years ago

Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs t

Misha Larkins [42]

Answer:

Speed greater than 4 m/s

Explanation:

Given that Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs to run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?

The distance = 12 m

Time = 3s

Speed = distance/time

Speed = 12/3

Speed = 4 m/s

Ms Kasper must run at speed more than 4m/s for her to save the cat.

3 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago