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Advocard [28]
3 years ago
5

The velocity v(t) of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s2)t - (6.2 m/s3)t2. What is the ave

rage acceleration of the particle between t = 1.0 s and t = 2.0 s (up to two significant figures) ?

A) -13 m/s2

B) -14 m/s2
C) 13 m/s2
D) 15 m/s2
E) None of the above
Physics
1 answer:
pantera1 [17]3 years ago
6 0

Answer:

average acceleration = −14.5 m/s²

so correct option is B) -14 m/s²

Explanation:

given data

v(t) = (2.3 m/s) + (4.1 m/s2)t - (6.2 m/s3)t²

t1 = 1.0 s

t2 = 2.0 s

to find out

What is the average acceleration

solution

we know that average rate of change on the interval is

average acceleration = \frac{Vf-Vi}{t2 - t1}    ............1

so here put value of t for final f and initial i

we get Vf =  2.3 + (4.1)(2) - (6.2)(2)²

and Vi = 2.3 + (4.1)(1) - (6.2)(1)²

and t2 - t1 = 2 - 1 = 1 s

put value in equation 1

average acceleration = [2.3 + (4.1)(2) - (6.2)(2)² ] - [ 2.3 + (4.1)(1) - (6.2)(1)² ]

average acceleration = −14.5 m/s²

so correct option is B) -14 m/s²

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a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi
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The radius of the sphere in meters is ,r = 10\sqrt{5} -20

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

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Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = \sqrt{5} -2

Hence, r = 10\sqrt{5} -20

So, the radius of the sphere in meters is ,r = 10\sqrt{5} -20

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5 0
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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
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