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3 years ago

8

A 25 kg rock resting on the bottom of a lake must be moved from the paths of boats. The rock has a density of 2350 kg/m^3. What

force is needed to lift the rock while under water?

1 answer:

algol133 years ago

7 0

Answer:

The force needed is the weight of the rock minus the buoyant force.

Explanation:

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Match the events related to the formation of the universe with the stages during which they occurred.

igomit [66]

Answer:

does this even make sense

Explanation:

8 0

3 years ago

Which of the following is not a way to make exercising on a cold day safer?

IRISSAK [1]

Answer:

D.

Explanation:

I think its D. because Tea during a cold workout isnt safe at all. Drinking tea while working out adds weight and you just simply dont need tea when working out.. Hope this helps man :)

4 0

3 years ago

Read 2 more answers

an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?

gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

$F_{net} = F_g - F = m\cdot g - F\\$

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

$1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N$

The tension force F in the supporting cables is 7245N

3 0

3 years ago

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and han

sleet_krkn [62]

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

4 0

3 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago