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1 year ago

Ok I am done waiting for an answer for 5 hours can somebody PLEASE PLEASE PLEASE help me TIMED ASSESMENT HERE!!!! PLEASE

Physics

2 answers:

Levart [38]1 year ago

7 0

Answer:

ACB

Explanation:

JulsSmile [24]1 year ago

3 0

Answer:

Explanation:

I'm going to assume that all the givens are in m/s. If I am correct, The orange line is speeding up. It is going from 4 m/s to 7 m/s. Its slant is from lower left to upper right. It is increasing in speed.

The green line is slowing down. It is going from 4 to 0 m/s. It's slant is from upper left to lower right.

The blue line is horizontal. It is neither slowing down or speeding up.

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All of the following processes are the ones that lead directly to the formation of sedimentary rock EXCEPT-

emmasim [6.3K]

Answer:

melting of rock deep underground.

Explanation:

The melting of rocks deep underground does not produce sedimentary rocks. Most igneous rocks are produced by this process.

When molten rocks underground called magma is solidified in the subsurface, it results into the formation of igneous bodies.

Sedimentary rocks forms by the accumulation of sediments.
Inside the basin where the sediments are deposited, they are compacted, cemented and lithified.
Chemical and physical weathering of rocks produces sediments.

4 0

2 years ago

Which subatomic particle has the least mass?

goblinko [34]

Answer:

electron

Explanation:

6 0

2 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

2 years ago

AYOO Hey plzzzzzzzzzzzzzzzzzzzzzzz

blagie [28]

Answer:

Hold on Ill answer it..When do u need it by

Explanation:

4 0

2 years ago

A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter

lutik1710 [3]

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72 m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion $(v_t=\omega\,R=\omega\,(0.25\,m)\,)$ - with $\omega$ being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals: $\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}$

Then the addition of these two velocities equals:

$\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}$

7 0

3 years ago