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kirill115 [55]
3 years ago
14

Explain how you would find the units for acceleration in a problem?

Physics
1 answer:
deff fn [24]3 years ago
6 0

Answer:

Calculating acceleration involves dividing velocity by time — or in terms of SI units, dividing the meter per second [m/s] by the second [s]. Dividing distance by time twice is the same as dividing distance by the square of time. Thus the SI unit of acceleration is the meter per second squared .

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A soccer ball of diameter d and mass m rolls up a hill without slipping, reaching a maximum height of h above the base of the hi
stich3 [128]

Answer:

Convert the units to standard units.

r = d / 2 (cm)

r = r in cm [ 1 m / 100 cm ]

m = m in g * [ 1 kg / 1000 g] = m in kg

Write the general equation for this problem.

The KE at the bottom of the hill = the PE at the top.

1/2 I w^2 + 1/2 m v^2 = mgh

I for a hollow sphere = 2/3xmxr^2

1/2 (2/3 x mr^2 x w^2) + 1/2 x m x v^2 = mgh

Now we need to relate w with v

v = w x r

w = v / r put this result in for w

1/2 (2/3 x mxr^2 z v^2 / r^2) + 1/2 m x v^2 = mgh

1/2 (2/3 x m v^2) + 1/2 x m x v^2 = mgh

1/3 m x v^2 + 1/2 mv^2 = mgh

5/6 m x v^2 = mgh Notice that the m's are all the same on both sides. They cancel.

5/6 v^2 = gx h

Solve for v

v^2 = 6/5 x g x h

Find the Rotational Kinetic Energy (KE_r)

PE - KE_translational = KE_r

mgh - 1/2 x m x v^2 = KE_r

Find the rotational rate at the bottom w

KE_r = 1/2 I x w^2

I = 2/3 m R^2

7 0
3 years ago
A sound wave is often called a pressure wave because there are regions of high and low pressure established in them medium throu
Bingel [31]

Explanation:

A sound wave is often called a pressure wave because there are regions of high and low pressure established in them medium through which the sound wave travels. The regions of high pressure are known as <u>Compressions</u> and the regions of low pressure are known as <u>Rarefactions</u> . Sound waves are composed of compressions and rarefactions. Compressions are the parts where the molecules are congusted and pressed together. However in the rarefactions molecules are relax and have enough space for expansion. Sound waves are the logitudnal waves and always been defined as the motion of the medium particles parallel to the wave motion.

7 0
3 years ago
A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the
vladimir1956 [14]

Answer:

b)  Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

3 0
3 years ago
Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their sep
olga55 [171]

The separation between them is \frac{r}{\sqrt{20} }

Concept :

If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.

Given,

Two point charges are brought closer together, increasing the force between them by a factor of 20.

Original force is

F = \frac{kq_{1} q_{2} }{r^{2} } -------- ( 1 )

Here, q_{1} , q_{2} are charges and r is the distance between them

New force F' = \frac{kq_{1q_{2} } }{r'^{2} } ----------- (2 )

Divide ( 1 ) and ( 2 )

\frac{F'}{F} = \frac{\frac{kq_{1}q_{2}  }{r'^{2} } }{\frac{kq_{1}_{2}  }{r^{2} } }

20 = \frac{r^{2} }{r'^{2} }

r' = \frac{r}{\sqrt{20} }

Given that force between them are increasing and therefore distance between them decrease by \frac{r}{\sqrt{20} }

Learn more about two point charges here : brainly.com/question/24206363

#SPJ4

8 0
2 years ago
According to Boyles’ law, PV = constant. If a graph is plotted with the pressure P against the volume V, the graph would be a(n)
Hunter-Best [27]

Answer:

C. hyperbola

Explanation:

From Boyle's law:

PV = k, where k is a constant

Solving for P:

P = k / V

At first glance, this equation doesn't fit any of the options.  But when you graph it, you can see that it's actually a <em>rotated</em> hyperbola.

4 0
3 years ago
Read 2 more answers
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