Help! pls, look at the attachment

What formal regions are located in the western hemisphere

Tju [1.3M]

Answer: Part of Europe, part of Africa, part of Antarctic and the entire America.

Explanation: Western hemisphere or west hemisphere encompasses all regions located west of the longitude 0 °, or meridian of Greenwich.

The formal regions located in the western hemisphere is Europe, Africa, Antarctic and America.

4 0

3 years ago

what is the acceleration produced by the resultant force acting on an object if the coefficient of friction acting between the b

LenKa [72]

The acceleration of the block is $-0.98 m/s^2$

Explanation:

The expression for the force of friction acting on the block is (assuming the surface is horizontal and flat):

$F_f = -\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the block

g is the acceleration of gravity

and where the negative sign means the direction of the force is opposite to that of the motion of the block

If this is the only force acting on the object, then this is also the resultant force, so we can rewrite Newton's second law as:

$F=ma\\-\mu mg = ma$

where

a is the acceleration of the block

Re-arranging the equation,

$a=-\mu g$

And so this is the expression for the acceleration of the block acted upon the force of friction.

In this problem, we have:

$\mu=0.1$

$g=9.8 m/s^2$

Solving,

$a=-(0.1)(9.8)=-0.98 m/s^2$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

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#LearnwithBrainly

5 0

3 years ago

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago

Suppose you sound a 1056-hertz tuning fork at the same time you strike a note on the piano and hear 2 beats/second. You tighten

Sindrei [870]

Answer:

The frequency of the piano string is either 1053 HZ or 1059 HZ.

Explanation:

Here we know that frequency of beats is equal to the difference between the frequencies between two waves .

Given that frequency of tuning fork is 1056 HZ .

Let the frequency of the piano be ' f ' .

Given that number of beats = 3.

We know that | 1056 - f | = 3 ;

⇒ 1056- f = ±3,

Upon solving this we get

f = 1056-3 and 1056 + 3

⇒ f = 1053 or 1059 .

6 0

3 years ago