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3 years ago

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An object with a charge of -3.2 uC and a mass of 1.0Ã—10^(-2) kg experiences an upward electric force, due to a uniform electric

field, equal in magnitude to its weight.
If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.

upward
downward
to the left
to the right

1 answer:

USPshnik [31]3 years ago

6 0

Answer:

The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.

Explanation:

The detailed solution can be found in the attachment below.

Thank you for reading and I hope this is helpful to you.

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Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area

yaroslaw [1]

Answer:

$5.63\cdot 10^{-6} C$

Explanation:

The capacitor of a parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

A is the area of each plate

d is the separation between the plates

$\epsilon_0$ is the vacuum permittivity

The energy stored in a capacitor instead is given by

$U=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

$U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}$

And re-arranging it

$Q=\sqrt{\frac{2U\epsilon_0 A}{d}}$

Now if we substitute

$d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J$

We find the charge stored on the capacitor:

$Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C$

7 0

3 years ago

The stopping distances associated with slower speeds approximate the forward visibilities provided by low beam lights. However,

aleksandr82 [10.1K]

Answer:

a. True

Explanation:

Illumination distance is the distance, up to which the light of the vehicle can reach. Hence, it is a maximum distance from the, that driver can see.

Stopping distance is the minimum distance required by the car to stop after brakes are applied.

So, in order to avoid any accident the illumination distance must be greater than the stopping distance. So, the driver can stop the vehicle in time, when he sees something in front of it.

Since, the stopping distance in this case is two or three times longer than illumination distance. Therefore, low beam light does not provide enough visibility in high speed driving situations.

Hence, the correct option is:

<u>a. True</u>

<u></u>

5 0

3 years ago

Which of the following could be used to sterilize plastic petri plates in a plastic wrapper? A) ultraviolet radiation B) microwa

Ivahew [28]

Gamma radiation could be used to sterilize plastic petri plates in a plastic wrapper.

Answer: Option E

<u>Explanation: </u>

It is known that radioactive radiations have greater ionizing powers compared to other electromagnetic radiations. Among the radioactive radiations, gamma rays are the most ionizing radiation.

So they are widely used in sterilizing process. As the penetration power of gamma is more compared to alpha and beta, in order to kill microorganisms gamma radiations are used.

When plastic petri plates wrapped with plastic are bombarded or sterilizing with gamma rays, the rays will ionise the micro-organism. Thus mutilating their DNA and preventing their reproduction.

8 0

3 years ago

Which year was pluto no longer considered a planet?.

katrin2010 [14]

2006, i hope this helps

6 0

2 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

4 years ago