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Elenna [48]
3 years ago
13

Compute your average velocity in the following two cases: (a) You walk 50.2 m at a speed of 2.21 m/s and then run 50.2 m at a sp

eed of 4.11 m/s along a straight track. (b) You walk for 1.00 min at a speed of 2.21 m/s and then run for 1.16 min at 4.11 m/s along a straight track.
Physics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

a) 2.87 m/s

b) 3.23 m/s

Explanation:

The avergare velocity can be found dividing the length traveled d by the total time t.

a)

For the first part we easily know the total traveled length which is:

d = 50.2 m + 50.2 m = 100.4 m

The time can be found dividing the distance by the velocity:

t1 = 50.2 m / 2.21 m/s = 22.7149 s

t2 = 50.2 m / 4.11 m/s = 12.2141 s

t = t1 +t2 = 34.9290 s

Therefore, the average velocity is:

v = d/t =2.87 m/s

b)

Here we can easily know the total time:

t = 1 min + 1.16 min = 129.6 s

Now the distance wil be found multiplying each velocity by the time it has travelled:

d1 = 2.21 m/s * 60 s = 132.6 m

d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m

d = 418.656 m

Therefore, the average velocity is:

v = d/t =3.23 m/s

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Answer:

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6 0
2 years ago
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A projectile is fired vertically upwards and reaches a height of 78.4 m. Find the velocity of projection and the time it takes t
Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2

Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

V^2 = U^2 - 2gH

g will be negative as the object is moving against the gravity

0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

U = 39.2 m/s

The time it takes to reach its highest point can be calculated by using the formula;

V = U - gt

Where V = 0

Substitute U and t into the formula

0 = 39.2 - 9.8 × t

9.8t = 39.2

t = 39.2/9.8

t = 4 seconds.

7 0
3 years ago
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
Young's Modulus refers to changes in the: a- Volume b- Length c- Body layers
bekas [8.4K]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

Stress is defined as force per unit area and strain is defined as proportional deformation in a material.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight

l = length of wire

A = area of cross section

\Delta l = change in length

Hence, the correct answer is Option b.

6 0
3 years ago
3 cases where kinetic energy become potential energy
finlep [7]
--  Toss a rock straight up.  The kinetic energy you give it
with your hand becomes potential energy as it rises. 
Eventually, when its kinetic energy is completely changed
to potential energy, it stops rising.

--  When you're riding your bike and going really fast, you come
to the bottom of a hill.  You stop pedaling, and coast up the hill.
As your kinetic energy changes to potential energy, you coast
slower and slower.  Eventually, your energy is all potential, and
you stop coasting.

--  A little kid on a swing at the park.  The swing is going really fast
at the bottom of the arc, and then it starts rising.  As it rises, the
kinetic energy changes into potential energy, more and more as it
swings higher and higher.  Eventually it reaches a point where its
energy is all potential; then it stops rising, and begins falling again.
8 0
3 years ago
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