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Novay_Z [31]
3 years ago
9

Projectile motion 1 and 2​

Physics
1 answer:
shtirl [24]3 years ago
8 0

1. Vf =62.57778m/s

h=62.5778

theta final=45°????

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Starting circuit One battery. 2 light bulbs in parallel; switch What is the voltage across the battery? What is the voltage acro
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Answer:

The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Explanation:

In a parallel circuit, the Voltage is same across all the components of the circuit and the current flowing through each component is added to get the total current across the circuit.

Let us say, the voltage across the circuit is V. The voltage across light bulb 1 and light bulb 2 is the the same i.e V

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Three plant like organisms which cannot produce their own food are
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a 70 kg desk sits on the floor motionless If gravity is pulling it with an acceleration of 9.8 m/s/s how much force is gravity e
adell [148]

Since force is mass*acceleration,

F = 70kg * 9.8 m/s

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3 years ago
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta
rusak2 [61]
From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s

The formula for radial acceleration is:
a = rω^2

So,
a = 53.2 (0.987) = 52.494 /s^2

The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2

If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
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7 0
3 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
3 years ago
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