A farmhand attaches a 25-kg bale of hay to one end of a rope passing over a

The superheroine Xanaxa, who has a mass of 68.1 kg , is pursuing the 75.3 kg archvillain Lexlax. She leaps from the ground to th

Alexandra [31]

The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

$U = mg\Delta h$

Here,

$\Delta h$ = Change in height

m = mass of super heroine

g = Acceleration due to gravity

The change in height will be,

$\Delta h = h_f - h_i$

The final position of the heroin is below the ground level,

$h_f = -16.1m$

The initial height will be the zero point of our system of reference,

$\Delta h = -16.1m-0m$

$\Delta h = -16.1m$

Replacing all this values we have,

$U = mg\Delta h$

$U = (68.1kg)(9.8m/s^2)(-16.1m)$

$U = -10744.81J$

Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J

4 0

3 years ago

(100 pts)<br>Why can't you use distance divide by time to calculate the instantaneous<br>speed?

Fiesta28 [93]

Answer:

Instantaneous speed means speed at any instant

that means Speed is changing with time

You know speed is distance/time

So that means distance is also changing with time

So we take infinitesimal small distance per infinitesimal small time As we assume speed is constant in infinitesimal small time dt

So, we take speed = ds/dt

ds = infinitesimal small distance

dt = infinitesimal small time

As its ratio is equal to speed at any instant

Note : We are taking infinitesimal small distance

But :) we are taking infinitesimal small time also

As you know if denominator is small fraction is large So fraction always give large value

So it's not O ( this makes confuse to most of students)

So, thanks

Good question

Keep thinking like this :)

4 0

3 years ago

Read 2 more answers

U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.

ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2, the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

5 0

2 years ago

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

denis23 [38]

The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.

5 0

3 years ago

You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time

7nadin3 [17]

When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.

Remember that this law tells us that

$F= ma$

$F= m \frac{\Delta v}{t}$

Therefore the best strategy is A. keep pushing the box forward at a steady speed

7 0

3 years ago