What is the abbreviation for mojave desert?

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If th

tensa zangetsu [6.8K]

Answer:

The net charge is $67.89 \mu C$

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = $12.0\times 10^{-3}\ kg$

Magnitude of electric field, E = $10^{3}\ N/C$

Angle made by the string, $\theta = 30^{\circ}$

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, $T = Tcos\theta$

Horizontal component of the tension in the string, $T = Tsin\theta$

Now,

Balancing the forces in the x-direction:

$Tsin\theta = QE$

$Q = {Tsin\theta}{E}$ (1)

Balancing the forces in the y-direction:

$Tcos\theta = mg$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

Thus

$T = \frac{mg}{cos\theta }$

$T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N$

Use T = 0.1357 N in eqn (1):

$Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C$

$Q = 67.89\times 10^{- 5}\ C = 67.89\mu C$

7 0

3 years ago

A 0.413 kg block requires 1.09 N

Virty [35]

Answer:

Explanation:

The equation for this is

f = μ $F_n$ where f is the frictional force the block needs to overcome, μ is the coefficient of static friction, and $F_n = w=mg$ (that means that the normal force is the same as the weight of the block which has an equation of weight = mass times the pull of gravity). Filling in:

1.09 = μ(.413)(9.8) and

μ = $\frac{1.09}{(.413)(9.8)}$ so

μ = .27

5 0

2 years ago

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) A baseball is throw

wel

Answer:8.1 m

Explanation:

Given

ball is launched from height of 3 m

initial velocity $u=10 m/s$

considering the ball is thrown vertically upward

Using $v^2-u^2=2 as$

where,

u=initial velocity

v=final Velocity

a=acceleration

s=distance

At maximum height final velocity will be zero

$v^2-u^2=2 gs$

$0-10^2=2(-9.8)\cdot h$

$h=5.10 m$

Therefore maximum height w.r.t ground is $3+5.10=8.10 m$

8 0

2 years ago

Suppose a skydiver jumps out of a plane at 15,000 meters above the ground. It takes him 2.0 seconds to pull the cord to deploy t

Evgesh-ka [11]

Answer:

The time he can wait to pull the cord is 41.3 s

Explanation:

The equation for the height of the skydiver at a time "t" is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.

When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 m = 15000 m - 4.9 m/s² · t²

-15000 m / -4.9 m/s² = t²

t = 55.3 s

Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.

6 0

2 years ago

A car is traveling at vx = 30 m/s . The driver applies the brakes and the car decelerates at ax = -5.0 m/s2 . What is the stoppi

____ [38]

Answer:

90 m

Explanation:

We use an ecuation of uniformly accelerated motion, which allows us to find the distance traveled by the car from the moment that driver applies the brakes until it stops completely, that is, when its final speed is zero.

$v_{f}^2=v_{o}^2+2ad$

We isolate the variable d, knowing that the final speed $v_{f}$ is zero

$d=\frac{-v_{o}^2}{2a}\\d=\frac{-(30m/s)^2}{2(-5m/s^2)}=90 m$

6 0

3 years ago