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Jlenok [28]
4 years ago
14

A 0.45 kg rock is projected from the edge of the top of a building with an initial velocity of 11.1m/s at an angle of 50 degrees

above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 16.9m from the base of the building. How tall is the building?
Physics
1 answer:
Misha Larkins [42]4 years ago
4 0

Answer:

bro just go to an easier class

Explanation:

#nerd

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SVETLANKA909090 [29]

Answer:

1. A

2. B

Explanation:

1. As long as The force is equally distributed the object will not move .

2. If the force in one side is greater than that of the other the object will move

3 0
3 years ago
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A
Pani-rosa [81]
The answer would be center of mass, B
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Name two factors that can affect the function of an enzyme
myrzilka [38]

Answer:

1. pH

2. Temperature

Hope this helps

3 0
3 years ago
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0
4 years ago
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

Explanation:

Since R(t)=\frac{V}{I(t)}, we calculate the resistance rate by deriving this formula with respect to time:

\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})

Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega

8 0
3 years ago
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