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aleksandr82 [10.1K]

3 years ago

5

A thin stream of water flows vertically downward. the stream bends toward a positively charged object when it is placed near it.

the positively charged object is then removed. what will happen to the same stream of water when a negatively charged object is placed near it? explain.

1 answer:

vfiekz [6]3 years ago

5 0

I dont know the answer i just want brainy points<span />

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A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

$\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .....(I)

Where, $\delta t'$ = proper time

$\delta t$ = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

$\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}$

For the second interval,

Put the value of $\Delta t'$ in the equation (I)

$\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value in the equation

$\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}$

$\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}$

$\Delta t_{2}=-5.77\times10^{8}\ s$

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

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3 years ago

Timers at a swim meet used four different clocks to time an event which recorded time is the most precise

tatiyna

The answer is b 55.2 that is correct

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frosja888 [35]

The change of state that is occurring is from the liquid state to the solid state.

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While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?

timurjin [86]

You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).

5 0

3 years ago