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grigory [225]
3 years ago
10

Please help I need this fast

Physics
2 answers:
STALIN [3.7K]3 years ago
8 0

Answer:

0 m/sec

Explanation:

b/c they were at rest and initial means at rest ,at rest means 0 HOPE THIS HELPS

vredina [299]3 years ago
7 0
I’m pretty sure 0 m/secs is right, sorry if I’m wrong
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Read the article Watering Livestock with Renewable Energy.
DaniilM [7]

Answer:

D. Wind and solar energy help save money and reduce air pollution.

Explanation:

7 0
3 years ago
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Can any one answer these two
dalvyx [7]

) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

7 0
3 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Scrat [10]

Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

E_k = E_p

mv^2/2 = mgh

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

v^2 = 2gh = 2*9.81*0.35 = 6.867

v = \sqrt{6.867} = 2.62 m/s

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

\omega = v / r = 2.62 / 0.175 = 14.97 rad/s

Where r is the rotation radius, or the distance from pivot point to the center of mass

v_0 = \omega R = 14.97*0.35 = 5.24 m/s

Where R is the distance from the pivot to the bottom end of the rod

5 0
3 years ago
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten
lions [1.4K]

Answer:

 h = 20 m

Explanation:

given.

height, h = 10 m

Potential energy at 10 m = 50 J

Kinetic energy at 10 m = 50 J

maximum height the ball will reach, H = ?

Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

P E = m g h

50 = m g x 10

mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

now, TE = Potential energy

 100 = m g h

5 h = 100

 h = 20 m

hence, the maximum height reached by the ball is equal to 20 m.

5 0
3 years ago
The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?
SIZIF [17.4K]
The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
I(r)= \frac{1}{r^2}
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put 
I= \frac{1}{r^2}= \frac{1}{(1)^2}=1
The intensity at r=3 m will be
I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}
Therefore, the sound intensity has decreased by a factor 1/9.
8 0
3 years ago
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