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lora16 [44]
2 years ago
13

Astronomers have discovered several volcanoes on io, a moon of jupiter. one of them, named loki, ejects lava to a maximum height

of 2.00 â 105 m. suppose another volcano on a different moon ejected lava at a height of 1.89 â 105 m where the acceleration of gravity is 1.72 m/s2.
Physics
1 answer:
r-ruslan [8.4K]2 years ago
7 0
The question seems to be incomplete. However, I can think of a possible logical question this problem could have. The equation for the maximum height attained by any object thrown upwards is:

H = v²/2g

I think the question would be determining the gravity in Io assuming that the initial velocity of the lava is the same. Then, the solution is as follows:

Let's use the other volcano to find v.
1.89×10⁵ m = v²/2(1.72 m/s²)
Solving for v,
v = 806.325 m/s

So, we use this to find g in Io.
2×10⁵ m = (806.325)²/2(g)
Solving for g,
<em>g = 1.6254 m/s²</em>
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a bobsled has a momentum of 1500 kg*m/s to the south. Air resistance reduces its momentum to 750 kg*m/s to the south. What impul
kherson [118]
The impulse is exactly the change in momentum,
and they have the same units.

The impulse imparted to the bobsled by the air
resistance is

                       (750 - 1500) = -750 kg-m/s.

It's negative because it reduced the bobsled's momentum.

If the impulse had come from a rocket engine on the back of
the bobsled, then it would have been a positive impulse.
7 0
2 years ago
Read 2 more answers
A person steps off the end of a 3.45 � high diving board and drops to the water below. (a) How long does it take for the person
777dan777 [17]

Answer:

(a) 0.84 s

(b) 8.22 m/s

(c) 15.33 m/s

Explanation:

u = 0 , h = 3.45 m, g = 9.8 m/s^2

Let time taken to reach the water is t and the velocity at the time of hitting of water surface is v.

(a) Use first equation of motion

s = u t + 1/2 a t^2

3.45 = 0 + 0.5 x 9.8 x t^2

t = 0.84 second

(b) Use first equation of motion

v = u + a t

v = 0 + 9.8 x 0.84 = 8.22 m/s

(c) If h = 12 m

use third equation of motion

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 9.8 x 12

v = 15.33 m / s

6 0
2 years ago
In an electricity experiment, a 1.0g plastic ball is suspended on a60-cm-long string and given an electric charge. A charged rod
alisha [4.7K]

a) Magnitude of the electric force: 3.57\cdot 10^{-3} N

b) Tension in the string: 0.010 N

Explanation:

a)

When the charged rod is brought near the ball, then the ball remains "suspended" in an inclined position. Therefore, we can analzye the forces acting in two perpendicular directions:

- Along the horizontal direction, we have the electric force F_E, pushing in one direction, and the component of the tension in the string acting in the opposite direction, T sin \theta, where T is the tension and \theta=20^{\circ} is the angle with the vertical

- Along the vertical direction, we have the weight of the ball, mg, acting downward (where m=1.0 g = 0.001 kg is the mass of the ball and g=9.8 m/s^2 is the acceleration due to gravity), and the component of the tension acting in the upward direction, T cos \theta

Therefore, since the ball is in equilibrium, we have the two equations:

T sin \theta =F_E\\Tcos \theta = mg

By dividing the two equations, we get

tan \theta=\frac{F_E}{mg}

an solving for the electric force, we find

F_E=mg tan \theta=(0.001)(9.8)tan 20^{\circ}=3.57\cdot 10^{-3} N

b)

The tension in the string can now be found by using either of the two equations above; for instance, by using the equation along the horizontal direction,

T sin \theta =F_E

Where

F_E=3.57\cdot 10^{-3} N is the electric force

\theta=20^{\circ} is the angle with the vertical

We find the tension in the string:

T=\frac{F_E}{sin \theta}=\frac{3.57\cdot 10^{-3} N}{sin 20^{\circ}}=0.010 N

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
2 years ago
You probably think of wet surfaces as being slippery. Surprisingly, the opposite is true for human skin, as you can demonstrate
Goshia [24]

Answer:

(A) 108 N

(B) 560 N

Explanation:

From the law of solid friction, for a body in contact with a flat surface,

F = \mu R

<em>F</em> is the maximum frictional force between the body and surface, <em>μ</em> is the coefficient of friction and <em>R</em> is the reaction between the body and the surface.

The maximum frictional force is the maximum weight of the rod the man can hold without slipping.

(A) His hands are dry. Then <em>μ</em> = 0.27.

<em>R</em> is the reaction of the rod against his hands. Because he presses on the rod perpendicular to the surface of the vertical rod, the reaction is the same as his grip force.

<em>R</em> = 400 N

Then, <em>F</em> = 0.27 × 400 N = 108 N

(A) His hands are wet. Then <em>μ</em> = 1.4.

<em>R</em> = 400 N

Then, <em>F</em> = 1.4 × 400 N = 560 N

7 0
2 years ago
Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the flo
inn [45]

Answer;

The pond's water level will fall.

Explanation;

Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.

When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.

Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.

When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.

Hope this Helps!!!

3 0
2 years ago
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