A group of physics teachers is taking some time off for a little putt-putt golf. The 15th hold has a large metal rim that putters must use to guide their ball towards the hole. Mr S guides a golf ball around the metal rim. When the ball leaves the rim, which path (1,2 & 3) will the golf ball follow and why?

Answers

Answer 1
Answer:
Call me nuts, but I can't shake the feeling that there's some kind of a
picture or drawing that's part of this question.  Unless I miss my guess,
you copied part of the question, and then decided to stop right there and
not let us see the picture.  If my hunch plays out, then we NEED to see
the picture in order to answer the question, and since you haven't shared
it, there is no way to write an answer.


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Help with #3 please

Answers

Answer:

The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²

Explanation:

Data

Ta = 0.125 s

Tb = 0.08 s

Δtab = 0.5 s

distance = 1 cm

Process

1.- Calculate va

va = 1/0.125 = 8 cm/s

vb = 1/0.08 = 12.5 cm/s

2.- Calculate Δv

Δv = 12.5 - 8

Δv = 4.5 cm/s

3.- Calculate acceleration

a = Δv / Δt

a = 4.5/0.5

a = 9 cm/s²

Given a 3.00 μF capacitor, a 7.75 μF capacitor, and a 5.00 V battery, find the charge on each capacitor if you connect them in the following ways. (a) in series across the battery 3.00 μF capacitor μC 7.75 μF capacitor μC (b) in parallel across the battery 3.00 μF capacitor μC 7.75 μF capacitor μC.

Answers

Answer:

a) Q1= Q2= 11.75×10^-6Coulombs

b) Q1 =15×10^-6coulombs

Q2 = 38.75×10^-6coulombs

Explanation:

a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as

1/Ct = 1/C1 + 1/C2

Given C1 = 3.00 μF C2 = 7.75μF

1/Ct = 1/3+1/7.73

1/Ct = 0.333+ 0.129

1/Ct = 0.462

Ct = 1/0.462

Ct = 2.35μF

V = 5.00Volts

To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance

Q = 2.35×10^-6× 5

Q = 11.75×10^-6Coulombs

Since same charge flows through a series connected capacitors, therefore Q1= Q2=

11.75×10^-6Coulombs

b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2

C = 3.00 μF + 7.75 μF

C = 10.75 μF

For 3.00 μF capacitance, the charge on it will be Q1 = C1V

Q1 = 3×10^-6 × 5

Q1 =15×10^-6coulombs

For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5

Q2 = 38.75×10^-6coulombs

Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.

A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?How can I get the potential energy ?

Answers

We know that -

P.E=m*g*h

Where,

m = mass

g = acceleration due to gravity

h=height

First we convert height into meters.

1 km = 1000 meters

3 km = 1000 * 3 meters = 3000 meters

So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-

P.E.= 2*3000*9.8

P.E.= 58800 Joules

P.E.= 58.8 kJ

Sairah's work is correct as far as it goes.  The potential energy of the ball
relative to the bottom of the hill is 58,800 Joules.

To address the second part of the question:
In order to get ahold of that energy, the ball must be returned to the bottom
of the hill. The most efficient way would be to drop it, so that it wouldn't have to
scrape along the grass on the way down. But that can only work if there's a sheer
cliff on one side of the hill. Otherwise, you just have to roll it down, and accept the
fact that it loses some of its energy to friction on the way.

However the ball gets to the bottom, the energy it has left shows up in the form
of kinetic energy, and 58,800 joules is a lot of kinetic energy.  If somehow the ball
could arrive at the bottom with ALL the energy it had at the top, it would be moving
at something like 540 miles per hour !

An elevator weighing 6,000 N moves up a distance of 10.0 meters in 30.0 seconds. a. How much work did the elevator’s motor do? b. What was the power of the elevator’s motor in watt and in horsepower?

Answers

Assume that work is equal to force x distance or w=fd. The force here is 6,000n and distance is 10.0 so therefore w=(6000n)(10.0m) and use that to do p=wt (power = work x time

Which of the following is NOT an early sign of heat illness?Dizziness
Headache
Flushed face
Excessive sweating
Struggling to perform

Answers

Struggling to perform is not an early sign of heat illnesses.

There are times when the body system will be unable to dissipate heat effectively. When this happens, heat illness bounds to occur.

Also, heat illnesses tend to occur when the balance of salt and water in the body becomes unbalanced, the temperature rises unexpectedly or the sweet couldn't keep an individual cool.

Some of the early signs includes Headache, dizziness, Muscle cramps, weak pulse among other related symptoms.

Based on the above explanation, we can see that struggling to perform is not an early sign of heat illnesses.

Learn more at: brainly.com/question/19609187

The answer is struggling to perform

If two solutions have unequal concentrations of a solute, the solution with the lower concentration is called

Answers

obviouskly lower concentrstion