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Ad libitum [116K]
3 years ago
14

Calculate 8 ∙ 10^-4 divided by 2 ∙ 10^2.

Physics
1 answer:
Digiron [165]3 years ago
7 0
8 ∙ 10^-4 / 2 ∙ 10^2 = (8/2) ∙ ((10^-4)*(10^-2)) = <span>4 ∙ 10^-6</span>
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If the measured resistance of a lamp is 45 ohms when it operates at a power of 80.0 watts, what is the current in the lamp?
melisa1 [442]
P= R.I^2 so 80=45*I^2
I=sqrt(80/45)= 1.33A
4 0
3 years ago
The ocean’s tides are not only affected by the shape of the coastlines, or slope of the ocean floor, but also by the gravitation
dexar [7]

Answer:

a) According to Newton's law of gravitation, as the distance between the Moon and the Earth decreases, the gravitational attraction increases and vice versa

The gravitational force of the Moon on the Earth causes the Earth to be slightly bulged on the side directly facing the Moon

The gravitational force also pulls the water bodies on the Earth's surface towards the Moon in the same manner and the effect is more pronounced due to the ability of the liquid water to assume a shape based on the magnitude of the gravitational field attracting it

Therefore, the region where the Moon is closest to the Earth we have a high tide as the water level rises and the region which is perpendicular to where the Moon is located has a  low tide

b) The two special types of tides are

1) The neap tide

2) The spring tide

Neap tide

Neap tide occurs when the Sun and Moon are 90° apart from each other when they are viewed by an observer from Earth

The gravitational pull of the Sun cancels (partially) the effect of the gravitational pull and tidal force of the Moon, resulting in minimum tidal range

Spring Tide

Spring tide occurs when the Earth, the Moon, and the Sun are simultaneously inline, such that the Sun reinforces the gravitational pull and tidal force of the Moon, resulting in a maximum tidal range

Explanation:

4 0
3 years ago
Read 2 more answers
A 200 kg object is initially at rest on a horizontal frictionless surface. At time t = 0 , a horizontal force of 100 N applied t
s344n2d4d5 [400]

Answer:

b

Explanation:

5 0
3 years ago
The sides of a square baseball diamond are 90 feet long. When a player who is between second and third base is 60 feet from seco
Sedbober [7]

Answer:

d(L)/dt  =  6,96 ft/s

Explanation:

We have a right triangle where the hypotenuse (L) is the distance between the player and home plate, the legs are the line between third base and home plate ( 90 feet ) and distance between the player and third base (x)  (over the line between second and third base). So we can write

L²  =  (90)²  +  x²

Applying differentiation in relation to time, on both sides of the equation we have:

2*LdL/dt  =  0  + 2*x d(x)/dt     (2)

In this equation we know:

d(x)/ dt  =  22 feet/sec

x = 30 ft

We need to calculate  L when the player is at 30 feet from third base

Then

L²  =  (90)² + (30)²

L²  =  8100 + 900

L  = √9000

L =94,87 feet

Then we are in condition for calculate  d(L)/dt from the equation

2*Ld(L)/dt  =  0  + 2*x d(x)/dt  

2*94,87 * d(L)/dt  =  2* 30* 22     ⇒  189,74 d(L)/dt = 1320

d(L)/dt  =  1320/ 189,74

d(L)/dt  =  6,96 ft/s

7 0
3 years ago
A bullet having the mass of 110g is fired. If it's velocity is 50m/s, calculate the kinetic energy​
Papessa [141]

\text{Kinetic energy,}\\\\E_k = \dfrac  12 mv^2\\\\~~~~~=\dfrac 12 \cdot 110  \times 10^{-3} \cdot 50^2\\\\~~~~~=137.5~ \text{J}

7 0
2 years ago
Read 2 more answers
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