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3 years ago

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Physics

1 answer:

anastassius [24]3 years ago

3 0

Answer:

P = density (p) * g * h

P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2

since kg m / s^2 = Newtons

The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density

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What is the direction of propagation of the wave if E⃗ =Ei^,B⃗ =−Bj^. Express the direction of the propagation vector, P⃗ , as a

nataly862011 [7]

Answer with Explanation:

We are given that

Electric field vector= $\vec{E}=E\hat{i}$

Magnetic field vector= $\vec{B}=-B\hat{j}$

We have to find the direction of propagation of the wave.

We know that

The direction of propagation of wave= $\vec{E}\times \vec{B}$

The direction of propagation of wave= $\hat{i}\times (-\hat{j})=-\hat{k}$

Because ( $\hat{i}\times \hat{j}=\hat{k}$ )

The wave propagate in -z direction.

Therefore, the component of the vector is (0,0,-1).

Hence, the direction of the propagation of wave is - z direction and component of direction of propagation vector is (0,0,-1).

6 0

4 years ago

An electron is moving directly toward you in a horizontal path when it suddenly enters a uniform magnetic field that is either v

blagie [28]

Answer:

To your left

Explanation:

The direction of the force exerted on charged particle due to a magnetic field is given by the right-hand-rule, where:

- The index finger indicates the direction of motion of the electron

- the middle finger gives the direction of the magnetic field

- the thumb gives the direction of the force if the particle is positively charged - otherwise, the direction is reversed

in this case, we have an electron (so, a negatively charged particle):

- The direction of motion (index finger) is horizontal, toward you

- The electron begins to curve upward as it enters the field, so this means that the force exerted on the electrons is upward --> the thumb must point downward (because the electron is negatively charged)

- The index finger gives us the direction of the magnetic field: therefore, to your left.

3 0

4 years ago

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

$E=U=\frac{1}{2}kx^2$

where

k = 7.91 N/m is the spring constant

$x=5.08 cm = 0.0508 m$

Substituting,

$E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J$

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

$E+W_f=K$

where

$K_f=\frac{1}{2}mv^2$ is the kinetic energy of the ball, with

$m=5.38 g = 5.38\cdot 10^{-3} kg$ being the mass of the ball

v being the final speed

$W_f = -F_f d$ is the work done by friction (which is negative since the force of friction is opposite to the motion), where

$F_f = 0.0323 N$ is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

$W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J$

So, the kinetic energy of the ball as it leaves the cannon is

$K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s$

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

$U=\frac{1}{2}kx^2$

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

$W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J$