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2 years ago

7.

Physics

2 answers:

Usimov [2.4K]2 years ago

7 0

Answer:

Explanation:

Tpy6a [65]2 years ago

4 0

Answer:

Explanation:

Magma rising up from the mantle at a divergent boundary, I got this question and I got it right.

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i

drek231 [11]

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $\frac{28}{10}$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

8 0

3 years ago

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. Ass

Naddika [18.5K]

Answer:

L = 4.711 *10^{-6} kg m2/s

Explanation:

$i =\frac{ml^3}{3}$

$= \frac{0.00*.15^2}{3}$

=4.5*10^-5

angular velocity

$\omega = \frac{2\pi}{60}$

= 0.1047 rad/s

the angular momentum,

$L = I\omega$

$L = 4.5*10^{-5}* 0.1047 rad/s$

$L = 4.711 *10^{-6} kg m2/s$

3 0

2 years ago

Read 2 more answers

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago

A 12000 kg boat is moving 4.25 m/s. Its engine pushes 9200 N forward, but the current pushes back at 12,500 N. How much times do

Verizon [17]

Answer:

15.5 seconds

Explanation:

Apply Newton's second law:

∑F = ma

-12500 + 9200 = (12000) a

a = -0.275 m/s²

v = at + v₀

0 = (-0.275) t + 4.25

t = 15.5 s

It takes the boat 15.5 seconds to stop.

7 0

3 years ago

What are the thee major wind systems

astra-53 [7]

Trade winds, prevailing westerlies, polar easterlies

8 0

3 years ago