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allochka39001 [22]
3 years ago
7

When the leaves are __________________, they fall straight down.

Physics
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:Done doing there job, in the winter they have a break

Explanation:

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The main reason why bells are made up of metals instead of wood is because metal is much more dense than wood, meaning that it resonates at much stronger frequencies. Wood has far too much air in it to make loud noises when struck.
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Why would a flat sheet of paper and a wad of paper with the same mass not fall through the air at the same rate?
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The flat sheet of paper has more surface area than the crumpled ball
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A force of 150N at an angle of 60 degree to the horizontal to pull a box through a distance of 50m calculate the work done
Alexxandr [17]
  • Force=150N
  • Angle=60°
  • Displacement=50m

\boxed{\sf W=Fscos\Theta}

\\ \sf\longmapsto W=150(50)cos 60

\\ \sf\longmapsto W=7500\times \dfrac{1}{2}

\\ \sf\longmapsto W=3750J

6 0
2 years ago
The position x, in meters, of an object is given by the equation:
LenaWriter [7]

Answer:

The SI units of A, B and C are :

m,\ m/s\ and\ m/s^2                  

Explanation:

The position x, in meters, of an object is given by the equation:

x=A+Bt+Ct^2

Where

t is time in seconds

We know that the unit of x is meters, such that the units of A, Bt and Ct^2 must be meters. So,

  • A=m
  • bt=m

b=\dfrac{m}{s}=m/s

  • Ct^2=m

C=m/s^2

So, the SI units of A, B and C are :

m,\ m/s\ and\ m/s^2

So, the correct option is (B).

3 0
3 years ago
An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

4 0
2 years ago
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