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3 years ago

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A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with

1 answer:

ivanzaharov [21]3 years ago

8 0

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

$PE_{gravitational}=PE_{spring}$

$mgh=0.5kx^{2}$ where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring

$mgdsin\theta=0.5kx^{2}$ where \theta is the angle of inclination and d is the sliding distance

Making x the subject then

$x=\sqrt {\frac {2mgdsin\theta}{k}}$

Substituting the given values then

$x=\sqrt{\frac {2\times 11\times 9.81\times 3\times sin 38}{2.9\times 10^{4}}}= 0.117240716\approx 11.72 mm$

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

2 years ago

A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?

noname [10]

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration ( $g=9.8 m/s^2$ , acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

$g=9.8 m/s^2$

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s$

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

$v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s$

So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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2 years ago

If Mrs. Reichelt throws a chromebook, because it won't login correctly, with a force of 8N, and the chromebook accelerates at 5m

suter [353]

Answer:

1.6 kg

Step-by-step Solution:

Since Force = mass × acceleration we have:

F = 8N

a= 5 m/s^2

m = ?

By plugging the values above into F=ma we obtain:

$F=ma\\\\8=m(5)\\\\\frac{8}{5}=\frac{m(5)}{5}\\\\m=\frac{8}{5}=1.6$

Therefore, the Chromebook has a mass of 1.6 kilograms.

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