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Rashid [163]
3 years ago
5

3 types of acceleration

Physics
2 answers:
emmasim [6.3K]3 years ago
5 0
3 types of acceleration are changes in speed, direction, and both simultaneously.
Svetach [21]3 years ago
5 0
Three types of acceleration are speeding up, slowing down, and turning. I have attached a photo that can help explain.

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25. The best description of scissor is:
ioda

Answer:

c.

Explanation:

From the options provided the best description of scissor is that scissors helps to scrap things down. This is the main purpose of scissors, to cut things, and by doing so you are breaking up a larger object into smaller and more manageable scraps. We can get rid of the other options because scissors are made of metal and therefore not soft and are not necessarily dull since they need to be sharp to fulfill their purpose. Although a pair of scissors has two holes it is not necessarily the best description of it.

4 0
3 years ago
True or False. Electromagnetic waves travel fastest through a vacuum.
olya-2409 [2.1K]
This
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Electromagnetic waves travel fastest through vacuum
4 0
3 years ago
A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f
uysha [10]

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

\Sigma F=ma

Replacing the values:

mg\sin 40=ma

We may cancel out the mass:

g\sin 40=a

Using the gravity constant as 9.8 meters per square second:

9.8\frac{m}{s^2}\sin 40=a

Solving the operations:

6.3\frac{m}{s^2}=a

Therefore, the acceleration is 6.3 meters per square second.

8 0
1 year ago
8. What is the mass of an object if a force of 34 N produces an acceleration of 4 m/s/s?
coldgirl [10]

Answer:

<h3>The answer is 8.5 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

where

f is the force

a is the acceleration

So we have

m =  \frac{34}{4}  =  \frac{17}{2}  \\

We have the final answer as

<h3>8.5 kg</h3>

Hope this helps you

7 0
3 years ago
Read 2 more answers
So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

5 0
3 years ago
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