Ask question

Ask question

All categories

3 years ago

12

Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Thr

ee of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

1 answer:

Gwar [14]3 years ago

6 0

Answer:

7.2N/C

Explanation:

Pls see attached file

You might be interested in

An object has an acceleration of 18.0 m/s/s. If the net force was doubled and the mass were tripled then the new acceleration wo

kenny6666 [7]

Answer:

12.0 m/s²

Explanation:

Newton's second law:

F = ma

Solving for acceleration:

a = F/m

If force is doubled and mass is tripled:

a' = (2F)/(3m)

a' = ⅔ (F/m)

a' = ⅔ (18.0 m/s²)

a' = 12.0 m/s²

4 0

3 years ago

An object of mass 700700 kg is released from rest 20002000 m above the ground and allowed to fall under the influence of gravity

qwelly [4]

Answer:

59.503987 seconds

Explanation:

b = Proportionality constant = 50 Ns/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object = 700 kg

We have the equation of velocity

$v(t)=\dfrac{mg}{b}+\left(V_0-\dfrac{mg}{b}\right)e^{\dfrac{bt}{m}}$

The equation of motion

$x(t)=\dfrac{mg}{b}+\dfrac{m}{b}\left(V_0-\dfrac{mg}{b}\right)(1-e^{\dfrac{bt}{m}})$

$x(t)=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})$

when x(t)=2000

$2000=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})\\\Rightarrow 2000\times \:50=\frac{700\times \:9.81}{50}\times \:50+\frac{9.81}{50}\left(0-\frac{700\times \:9.81}{50}\right)\left(1-e^{\frac{50t}{700}}\right)\times \:50\\\Rightarrow 6867-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)=100000\\\Rightarrow 50\left(-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)\right)=93133\times \:50\\\Rightarrow \frac{-67365.27\left(1-e^{\frac{50t}{700}}\right)}{-67365.27}=\frac{4656650}{-67365.27}\\\Rightarrow 1-e^{\frac{50t}{700}}=-69.12538\dots\\\Rightarrow -e^{\frac{50t}{700}}=-70.12538\dots\\\Rightarrow t=14\ln \left(70.12538\dots \right)\\\Rightarrow t=59.50398\ s$

The time taken is 59.503987 seconds

8 0

3 years ago

Which actions are essential to active participation? Check all that apply.

erastova [34]

Answer:

2, 3, 4, 6.

Explanation:

8 0

3 years ago

Read 2 more answers

A worker on the roof of a house drops his 0.46 kg hammer, which slides down the roof at constant speed of 9.88 m/s. The roof mak

nekit [7.7K]

Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

The horizontal speed
the time the hammer takes to fall from the roof to the ground

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

$v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s$

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

$x(t)=8.8 m/s *t$

$y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}$

Now we calculate the time the hammer takes to get to the floor

$17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s$ or $t=-2.38s$

Now, we keep the positive time result and calculate the horizontal distance travelled

$x(1.47s)=8.8m/s*1.47s=12.94m$

3 0

3 years ago

The instruments carried by a spacecraft are called the

Sedaia [141]

I believe the answer is C- payload

6 0

2 years ago

Read 2 more answers