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2 years ago

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A car begins at rest at a stoplight. After the light turn green, the car begins to accelerate 5 m/s2 over the span of 3.6 second

s. What is the car’s displacement over this period of time?
Hint you can use this equation Xf = Xi + vi · t + 1/2 · a · t2

1 answer:

trasher [3.6K]2 years ago

5 0

Answer:

d=9m

Explanation:

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A stationary car with a mass of 1500 kg reaches a velocity of 15 meters per second, 5 seconds after starting. What is the car's

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Answer:

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3 years ago

A) Julie is running to the right at 5 m/s, as shown in the 1st figure. Balls 1 and 2 are thrown toward her at 10 m/s by friends

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Part 1)

here Julie is running at speed 5 m/s

So here two balls are thrown at speed 10 m/s towards Julie with respect to her friend standing on the ground.

So here this all speed is real speed of all.

Now as per Anita (let say she is one of her friend standing on ground) the speed of two balls will be same as the given speed as she is observing from ground or stationary frame

As per the frame of Julie

speed of ball 1

$v_{1j} = v_1 - v_j$

$v_{1j} = 10 - 5 = 5 m/s$ towards her in same side

speed of ball 2

$v_{2j} = v_2 - v_j$

$v_{2j} = -10 - 5 = -15 m/s$ towards her from opposite side

Part b)

Now in this case the speed of two balls is given with respect to Julie

so we can say

for ball 1

$v_{1j} = v_1 - v_j$

$10 = v_1 - 5$

so in ground frame speed of ball 1 is

$v_1 = 15 m/s$

Similarly for ball 2

$v_{2j} = v_2 - v_j$

$-10 = v_2 - 5$

So speed in ground frame of ball 2 is

$v_2 = - 5 m/s$

so its 5 m/s from opposite side

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3 years ago

A statue is made of bronze, which is an alloy of about 80% copper and 20% tin. Which metal is in the higher concentration in thi

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3 years ago

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3 years ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

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2 years ago