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Elza [17]
2 years ago
5

A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 m from the takeoff point. part a if the

kangaroo leaves the ground at a 21 ∘ angle, what is its takeoff speed
Physics
2 answers:
jeka942 years ago
8 0
<span>Each jump = 11 m
 Angle of the speed = 21 degrees.
Gravitational Acceleration g = 9.8 m/s^2
  Initial velocity V^2 = (Rg)/sin2A
 V^2 = 11 x 9.81 / sin (2 x 21)
 V^2 = 107. 91 / 0.669
 V^2 = 161.27
 V = 12.70
Now we got the velocity and calculate the horizontal angle, Velocity (horizontal) = vcosA = 12.70 x cos 21 Velocity = 11.85 m/s</span>
Svetllana [295]2 years ago
6 0
We have that the maximum rank of the kangaroo is given by:
 R = v0 ^ 2 sin (2θ) / g
 where,
 v0 = initial velocity
 θ = angle of the velocity vector formed from the horizontal
 g = gravity
 Clearing the speed we have:
 v0 ^ 2 = (R * g) / (sin (2θ))
 Substituting values
 v0 = root (((11) * (9.8)) / (sin (2 (21 * (pi / 180)))))
 v0 = 12.69 m / s
 answer
 its takeoff speed is 12.69 m / s
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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

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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

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At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

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Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

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2 years ago
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