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svetoff [14.1K]
2 years ago
9

Someone please help its a simple power problem.

Physics
1 answer:
SOVA2 [1]2 years ago
7 0
Well 200 doubled or (x2)=400 if that’s what it means
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You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

Explanation:

u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

m = Mass of each lump = 30\ \text{g}

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
3 years ago
You wrap a wire around a piece of iron. If you slowly increase the strength of an electric current flowing through the wire, you
Maksim231197 [3]
The correct answer is A. the magnet to become stronger

The stronger the electric current in the piece of metal, the stronger the magnetic field will be.
4 0
3 years ago
Read 2 more answers
URGENTE ¿Cuál de los siguientes términos corresponde al concepto presión? * A. Es la fuerza perpendicular que ejerce un cuerpo s
melomori [17]

Answer:

Respuesta correcta, opción D: Es la fuerza que un cuerpo ejerce perpendicularmente sobre el área en la que actúa.

Explanation:

La definición de presión es la fuerza que un cuerpo ejerce en dirección perpendicular sobre el área en la que actúa.

Cuando se aplica una fuerza sobre la superficie de un cuerpo, la presión es la siguiente:

P = \frac{F}{A}

En donde:

F es la fuerza aplicada.

A es el área del cuerpo.  

Por lo tanto la opción correcta es la D: es la fuerza que un cuerpo ejerce perpendicularmente sobre el área en la que actúa.

Espero que se sea de utilidad!      

6 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
8) Find the X and Y component of 10degree vector that has 5N.
Deffense [45]

Answer:

Fx  = 4.92 [N]

Fy = 0.868 [N]

Explanation:

Let's take the 10 degrees as a measure from the horizontal component to the vector.

Thus taking the components in the X & y axes respectively:

Fx = 5*cos(10) = 4.92 [N]

Fy = 5*sin(10) = 0.868 [N]

3 0
3 years ago
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