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3 years ago

9

Monarch butterflies can sustain air speeds of 4.8 m/s while migrating. A certain monarch wants to fly to a location due east 4.7

km away. A steady 2.4 m/s breeze is blowing from the west to the east. How long will it take the monarch to make the trip to the location to the east?

2 answers:

AnnyKZ [126]3 years ago

6 0

It will take the butterfly 2.3 km to reach the destination

wariber [46]3 years ago

4 0

That would probably be 2.3 km

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What is the independent variable? What is the dependent variable? Please help!

e-lub [12.9K]

graphic too wide.

indep variable is one you control

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2 years ago

True or false. Students with a Learners License may not receive a motorcycle endorsement.

viktelen [127]

This is a true statement

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2 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th

vladimir1956 [14]

Answer:

h = 22.35 m

Explanation:

given,

initial speed of the rock,u = 0 m/s

length of the window,l = 2.7 m

time taken to cross the window,t = 0.129 s

Speed of the rock when it crosses the window

$v = \dfrac{l}{t}$

$v = \dfrac{2.7}{0.129}$

v = 20.93 m/s

height of the building above the window

using equation of motion

v² = u² + 2 g h

20.93² = 0² + 2 x 9.8 x h

h = 22.35 m

Hence, the height of the building above the top of window is equal to h = 22.35 m

8 0

2 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

<em>so the distance between them is 11.5 m</em>

6 0

2 years ago

In a thundercloud there may be electric charges of 45.0 C near the top of the cloud and -45.0 C near the bottom of the cloud. Th

Gala2k [10]

Answer:

The electric force on the top charge is $F=3.44\times 10^6\ N$ .

Explanation:

Given that,

Electric charges in a thundercloud, $q_1=q_2=45\ C$

The distance between charges, d = 2.3 km = 2300 m

Let F is the electric force on the top charge. The electric force is given by the formula as :

$F=\dfrac{kq^2}{d^2}$

$F=\dfrac{9\times 10^9\times (45)^2}{(2300)^2}$

$F=3445179.58\ N$

or

$F=3.44\times 10^6\ N$

So, the electric force on the top charge is $F=3.44\times 10^6\ N$ . Hence, this is the required solution.

7 0

3 years ago