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2 years ago

What do these letters stand for P=mv

Physics

1 answer:

victus00 [196]2 years ago

4 0

Answer:

The equation for momentum of a piece of matter.

In either case, the momentum would be less than a linebacker hitting you at full speed. The equation for momentum is written: p = mv where p stands for momentum. That is, mass times velocity equals momentum.

Explanation:

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Tarzan and Jane. Because of your concern that incorrect science is being taught to children when they watch cartoons on TV, you

creativ13 [48]

Answer:

The maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

Explanation:

Let

m = Mass of Tarzan

M = Mass of Jane

Given

M = 2m

To calculate the maximum height Tarzan and Jane can swing, we make use of the potential energy at their initial and final position.

Reason being that;

At both the initial and final position, velocity is 0, so there's no kinetic energy.

And the potential energy remains the same (i.e constant) at any given point in the system.

Using P.E = mgh.

At initial position, PE1 = mgh

At final position, PE2 = (m + M)gH.

Where h and H represent the initial and final heights.

m + M is the new weight after Jane and Tarzan swing

Equating PE1 to PE2

mgh = (m + M)gH

By substituton (M = 2m)

mgh = (m + 2m)gH

mgh = 3mgH

Make H the subject of the formula

H = mgh/3mg

H = ⅓h

Hence, the maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

From the question, the new height looks to be about ½ that of Jane's original position; i.e. ½h

The calculated height is smaller than what the cartoon is showing;

We can conclude that the cartoon is wrong.

4 0

3 years ago

A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d

QveST [7]

Answer:

The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

(target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

(10.2 s)(1350 m/s) = 13,770 m

We are interested in the target movement, so we can solve for that:

(target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

-94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

The target's speed is found by dividing the distance it covers by the time it takes.

√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0

3 years ago

What value is closest to the mass of the atom?....

Sveta_85 [38]

The answer would be 6amu

6 0

3 years ago

Read 2 more answers

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

3 years ago

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w

Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0

3 years ago

Read 2 more answers