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vredina [299]
3 years ago
10

A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 7.1 m making 3.9 revolutions

every 9.4 second. What is the centripetal acceleration of the ball?
Physics
1 answer:
IgorC [24]3 years ago
7 0

Answer:

48,2 m/s²

Explanation:

We're gonna use the Centripetal Acceleration formula: v² / r but before that, we got to know the velocity, that is not shown clearly to us, so....

To know the velocity let's calculate the distance that the ball traveled

The circumference of a circle formula is:

2piR

2 . 3,14 . 7,1 | That is equal to 44,588 m

We know that the ball traveled this distance 3,9 times, so...

44,588 . 3,9 = 173,8932 m

Ok, now we have the distance, just need to know the time, that is 9.4 seconds.

Velocity = Distance / Time

V = 173,8932 / 9,4

V = 18,5 (approximate)

So...

We are back to the first formula:

Ca = v² / r

Ca = 18,5² / 7.1

Ca = 48,2 m/s² (approximate)

I hope it is correct, hahaha.

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Water boiling, no cheamical bonds have been altered.
7 0
3 years ago
A gas has a pressure of 48atm in a 15.5L container. It was found that at 25∘C the gas occupied a volume of 25L and had a pressur
EleoNora [17]

Answer:

130.165636364°C

Explanation:

P = Pressure

V = Volume

n = Number of moles

R = Gas constant = 0.082 L atm/mol K

From ideal gas law we have

PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles

PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K

The initial temperature is 403.315636364-273.15=130.165636364\ ^{\circ}C

3 0
4 years ago
What are the answers please help
aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.)  10.31 cm

e.)  10.24 cm  

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.)  If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm  

3 0
4 years ago
A crate is pushed up each of the two ramps shown in the diagram below. Based on the concept that simple machines make work easie
Naddik [55]

Answer:

ramp b requires less force than ramp a

Explanation:

7 0
3 years ago
Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing
Juliette [100K]

Answer: MR²

is the the moment of inertia  of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements  are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

3 0
3 years ago
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