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2 years ago

7

On her way home from work, Brenda drove 20 miles at 60 miles per hour. Due to poor weather conditions, she then reduced her spee

d to 30 miles per hour for the remainder of the trip. Her entire commute to work is 80 miles in distance. Based on the information provided above, which numerical value best represents how much less time would it have taken Brenda had she driven the entire distance at 40 miles per hour?
A. 1/12 hour
B. 1/6 hour
C. 1/4 hour
D. 1/3 hour
E. 2/3 hour

1 answer:

rjkz [21]2 years ago

3 0

Answer:

D

Explanation:

Speed = distance / time

her time for the first journey = 20 miles / 60 miles/hr = 1/3 hr

her time for second part of the journey = her remaining distance / her speed = (80 - 20) miles / 30 miles/hr = 60 miles / 30 miles/hr = 2 hrs

total time spend by her = 2 hr+ 1/3 hr = 2 1/3 hrs

her traveling the distance at 40 miles per hour = 80 miles / 40 miles /hr = 2 hrs

the time less she would drive if she drive the entire distance at 40 miles/hr = 2 1/3 hrs - 2 hrs = 1/3 hr

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Answer:

(a) 0.0178 Ω

(b) 3.4 A

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(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

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L = length of the wire = 6.7 m

Resistance of the wire is given as

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$R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }$

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

$i = \frac{V}{R}$

$i = \frac{0.060}{0.0178}$

i = 3.4 A

(c)

Current density is given as

$J = \frac{i}{A}$

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J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

$E = \frac{J}{\sigma }$

$E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}$

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

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(a) 18.9 m/s

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We can calculate the components of the initial velocity of the stone as it is launched from the ground:

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The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

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We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

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At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

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$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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Svetllana [295]

Answer:

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The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

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Answer:

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