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kykrilka [37]
3 years ago
11

Can someone explain this

Physics
2 answers:
Alborosie3 years ago
8 0

Answer:

gravity

Explanation:

That would make the most sense from what I know

stepladder [879]3 years ago
6 0
1) F= mass • acceleration
4000 N = 50g • acceleration
———— ——
50g 50g
a = 80m/s^2

a = v/t
v = at
v= 80(.01)
v = .8 m/s

2) The billeted speed decreases so quickly because it has reach its maximum speed.

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The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b
ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

8 0
3 years ago
Alex has to pick up a 20 N stack of documents and raise them 5 meters from the floor.
Serjik [45]

Answer:

1.Work done=100 joules

2.potential energy

3.kinetic energy

Explanation:

W=Force x distance

w= 20 N x 5 M

W = 100 JOULES

6 0
3 years ago
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
During a particular thunderstorm, the electric potential difference between a cloud and the ground is vcloud - vground = 3.50 10
wariber [46]
<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>
8 0
3 years ago
A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1
ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) =  1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0
2 years ago
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