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Anna71 [15]
2 years ago
14

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

aunch point. What was the arrow's initial speed? How long did it take for the arrow to first reach a height of 17.5 m above its launch point?
Physics
1 answer:
sdas [7]2 years ago
4 0

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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A snowboarder goes down the hill with a slope of 28° if friction acts on him as he slides down which of the following is the cor
xxMikexx [17]

Answer:

A

Explanation:

All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A

6 0
3 years ago
a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v
shutvik [7]
Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = e^{0.3098}


v₂  =  19* e^{0.3098}

v₂ = 25.8999

v₂ ≈ 26 L        Option b.
6 0
3 years ago
8. What is the frequency of green light waves that have a wavelength of 5.2 x 10-7 m.? The speed of light is 3.0 x 108 m/s
o-na [289]

Answer:

f=5.76\times 10^{14}\ Hz

Explanation:

We need to find the frequency of green light having wavelength o5.2\times 10^{-7}\ m. It can be calculated as follows :

c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{5.2\times 10^{-7}}\\\\f=5.76\times 10^{14}\ Hz

So, the required frequency of green light is equal to 5.76\times 10^{14}\ Hz.

4 0
3 years ago
A boy kicks a soccer ball, giving it an initial speed of 7.5m/s at an angle of 27° above the horizontal(=ground). How high will
Gelneren [198K]

When the initial speed given is 7.5m/s at an angle of 27° , ball will go

4.637 meters.

Assume no air opposition to the ball ;

Vertical component of ball is sin 27° =  0.453

0.453* 7.5 = 3.404 meters /sec

Time taken to reach ground is :

3.404 = -3.404+9.8*t

t= 6.808/9.8= 0.694 sec

Horizontal component is 7.5*cos27°= 6.682m/s

Distance = speed * time

=6.682 * 0.694

=4.637 meters

Horizontal distance it can cover in 0.694 sec is 4.637 meters

So range of ball is 4.637 meters.

Form of motion experienced by an object or particle that is projected near surface of the earth and moves along a curve is called Projectile motion. Three types of projectile motion are Horizontal projectile motion. Oblique projectile motion and Projectile motion on an inclined plane.

To know more about projectile motion, refer

brainly.com/question/24216590

#SPJ13

3 0
1 year ago
A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample
Musya8 [376]

Answer:

remain 1s60

Explanation:

I took away sample

7 0
2 years ago
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