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4 years ago

HELP ME !!!Question 4 of 10

Physics

2 answers:

Dmitrij [34]4 years ago

8 0

Answer:

The answers are C and D

Explanation:

Apex

kobusy [5.1K]4 years ago

8 0

Answer:

C: From the pot to the water

D: From the burner to the bottom of the pot

Explanation:

--APEX-- HOPE THIS HELPS!!

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A current of 3.60A flows for 15.3s through a conductor. Calculate the number of electrons that pass through a point in the condu

Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

(55.08 Coul) x (6.25 x 10^18 e/coul) = 3.4425 x 10^20 electrons

7 0

4 years ago

At t = 0, Ball 1 is dropped from the top of a 22 m-high building. At the same instant Ball 2 is thrown straight up from the base

Crazy boy [7]

Answer:

The two balls pass each other at a height of 5.53 m

vf1=17.97 m/s

vf2=-5.96 m/s

Explanation:

Vertical Motion

An object thrown from the ground at speed vo, is at a height y given by:

$y=vo.t-g.t^2/2$

Where t is the time and $g=9.8\ m/s^2$

Furthermore, an object dropped from a certain height h will fall a distance y, given by:

$y=g.t^2/2$

Thus, the height of this object above the ground is:

$H = h-g.t^2/2$

The question describes that ball 1 is dropped from a height of h=22 m. At the same time, ball 2 is thrown straight up with vo=12 m/s.

We want to find at what height both balls coincide. We'll do it by finding the time when it happens. We have written the equations for the height of both balls, we only have to equate them:

$vo.t-g.t^2/2=h-g.t^2/2$

Simplifying:

$vo.t=h$

Solving for t:

$t=h/vo=22/12=1.833\ s$

The height of ball 1 is:

$H = 22-9.8.(1.833)^2/2$

H = 5.53 m

The height of ball 2 is:

$y=12\cdot(1.833)-9.8\cdot(1.833)^2/2$

y=5.53 m

As required, both heights are the same.

The speed of the first ball is:

$vf1=g.t=9.8\cdot 1.833=17.97\ m/s$

vf1=17.97 m/s

The speed of the second ball is:

$vf2=vo-gt=12-9.8\cdot 1.833=-5.96\ m/s$

vf2=-5.96 m/s

This means the second ball is returning to the ground when both balls meet

3 0

3 years ago

A disk-shaped grindstone of mass 1.7 kg and radius 8 cm is spinning at 730 rev/min. After the power is shut off, a woman continu

Anastasy [175]

Answer:

0.186 N-m

Explanation:

mass of the grindstone, $m=1.7 kg$

radius, $r=8 cm$

Frequency, $f=730 rev/min = 12.16 rev/s$

time, $t=9s$

final angular velocity, $\omega=0$

Initial angular velocity,

$\omega_o=2\pi f\\=2\pi (12.16) rad/s\\= 76.36 rad/s$

Angular acceleration of the grind stone is:

$\alpha=\frac{\omega-\omega_o}{t}\\\Rightarrow \alpha =\frac{0-76.36}{9} = -8.48 rad/s^2$

Moment of inertia:

$I=mr^2+mr^2=2mr^2$

$I=2\times 1.7 kg\times (0.08m)^2= 0.022kg-m^2$

Torque exerted by the ax on the grind stone is:

$\tau=I\alpha\\\tau=0.022\times (-8.48) \\\tau=0.186N-m$

8 0

4 years ago

A cannon, elevated at 40∘ is fired at a wall 300 m away on level ground, as shown in the figure below. The initial speed of the

Gre4nikov [31]

Vo = 89 m/s
angle: 40°

=> Vox = Vo * cos 40° = 89 * cos 40°

=> Voy = Vo. sin 40° = 89 * sin 40°

x-movement: uniform => x =Vox * t = 89*cos(40)*t

x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s

y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2

y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.

7 0

3 years ago

Read 2 more answers

A cylinder of radius r=10.0 cm and height h=20.0 cm

Olenka [21]

D I think is the correct answer
If the cylinder is slightly

7 0

2 years ago