11) A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to

Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created is $P_{avg} = F * v_{avg}$

Explanation:

From the question we are told that

The that the average power is mathematically represented as

$P_{avg} = \frac{W }{\Delta t }$

Where W is is the Workdone which is mathematically represented as

$W = F * s$

Where F is the applies force and s is the displacement due to the force

So

$P_{avg} = \frac{F *s }{\Delta t }$

Now this displacement can be represented mathematically as

$s = v_{avg} * \Delta t$

Where $v_{avg }$ is the average velocity and $\Delta t$ is the time taken

So

$P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }$

=> $P_{avg} = F * v_{avg}$

3 0

2 years ago

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

2 years ago

Read 2 more answers

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

AleksandrR [38]

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

5 0

2 years ago

) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force

Nana76 [90]

Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others.
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
$F_x=F_3+F_2cos(45^{\circ})$
$F_y=F_1+F_2sin(45^{\circ})$
Let's figure out each component:
$F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\$
$F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}$
Total force acting on the charge would be:
$F=\sqrt{F_x^2+F_y^2}$
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
$F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N$
$F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N$
Now we can find the total force acting on a single charge:
$F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N$
As said before, intensity of the force acting on charges is the same for all of them.