All categories

2 years ago

A FORCE OF 3,200 N WEST is applied to a 160 kg mass. What is the acceleration of the mass

1 answer:

3 0

30098
. Hi I’m sorry to hear about the family and stuff but I’m still coming in for the day so I’m not going out to the beach today or just a little later lol lol I’m going on the beach with the girls lol I just need a shower lol I need a shower lol I’m just going on the

You might be interested in

Add the vectors:

Anettt [7]

Vector 1 has components

$x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m$

$y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m$

and vector 2 has

$x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m$

$y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m$

Add these vectors to get the resultant, which has components

$x_{\rm total}\approx11.133\,\mathrm m$

$y_{\rm total}\approx13.268\,\mathrm m$

The magnitude of the resultant is

$\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m$

with direction $\theta$ such that

$\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ$

or about 50º N of E.

8 0

3 years ago

a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a

Whitepunk [10]

Answer:

$W = -2.76\times 10^{-9}~J$

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

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

where C can be calculated using the plate separation and area.

$C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon$

Therefore, the potential energy in the first case is

$U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon$

In the second case:

$C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon$

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

$W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J$

6 0

3 years ago

All of the following are part of the electromagnetic spectrum but

Lera25 [3.4K]

D)sound waves the electromagnetic spectrum has to do with colors<span />

3 0

3 years ago

Read 2 more answers

Which best compares kinetic energy and temperature?

shutvik [7]

Answer:

the answer is c

Explanation:

because of Kinect energy being able to transfer form one particle to an other, while temperature allows the kinetic energy go higher in rate so we can say that in increase in temperature the kinetic energy will rise by this we can measure kinetic energy by temperature

3 0

3 years ago

a lemming running 2.87 m/s runs off a horizontal cliff. It lands in the water 5.32 m from the base of the cliff. How high was th

Vlad1618 [11]

Looking at the scenario, you can tell that this is a Type 1 Projectile, or a horizontal projectile. You would need a couple of things before you can solve this problem.

First you need to take a look at your given:

Vix = 2.87 m/s (Horizontal velocity)
dx = 5.32m (Horizontal distance)

What you need is your vertical distance or your height. The standard formula for projectile displacement is:
$d = vit + \frac{1}{2}gt^{2}$

Where:
vi = initial velocity
g = Acceleration due to gravity = 9.8m/s^2
t = time

To specifically get the vertical distance you just need to put in the y-components. Now remember that initially, for a horizontal projectile, there is not vertical movement, only forward. So your vi = om/s which will leave you with the equation:

$dy=viyt + \frac{1}{2}gt^{2}$
$dy=(0m/s)t + \frac{1}{2}gt^{2}$
$dy=\frac{1}{2}gt^{2}$

Your new formula would then be:
$dy=\frac{1}{2}gt^{2}$

But if you look at your problem, you can see that there is no time given. With what was given to you, you can solve it by deriving it from the x component formula:

$dx=vixt$

Plug in what you know and solve for what you don't know.
$dx=vixt$
$5.32m=(2.87m/s)t$
$\frac{5.32m}{2.87m/s}=t$
$1.85s=t$

Your time is then 1.85s.

Now that you know the time, you can now solve for your vertical distance/displacement:

$dy=\frac{1}{2}gt^{2}$
$dy=\frac{1}{2}(9.8m/s^{2}(1.85s)^{2}$
$dy=\frac{1}{2}(9.8m/s^{2}(1.85s)^{2}$
$dy=16.77m$

The cliff was then 16.77m high.

5 0

3 years ago