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const2013 [10]
3 years ago
11

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

irst one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 m/s. The first one is thrown at an angle of 75.0° with respect to the horizontal. How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?
Physics
1 answer:
marin [14]3 years ago
3 0

Answer:

3.60342 seconds

Explanation:

v = Initial velocity of snowball = 25 m/s

g = Acceleration due to gravity = 9.81 m/s²

\theta_1 = First angle = 75°

The second angle will be

\theta_2=90-\theta_1=90-75=15^{\circ}

Horizontal speed for first throw

v_{x1}=vcos\theta_1\\\Rightarrow v_{x1}=25\times cos75\\\Rightarrow v_{x1}=6.47047\ m/s

Horizontal speed for second throw

v_{x2}=vcos\theta_1\\\Rightarrow v_{x2}=25\times cos15\\\Rightarrow v_{x2}=24.14814\ m/s

Horizontal range is given by

R=\frac{v^2sin(2\theta)}{g}\\\Rightarrow R=\frac{25^2\times sin(2\times 75)}{9.81}\\\Rightarrow R=31.85\ m

Time period for first throw

t_1=\frac{R}{v_{x1}}\\\Rightarrow t_1=\frac{31.85}{6.47047}\\\Rightarrow t_1=4.92236\ s

Time period for second throw

t_2=\frac{R}{v_{x1}}\\\Rightarrow t_2=\frac{31.85}{24.14814}\\\Rightarrow t_2=1.31894\ s

The time difference is

t=4.92236-1.31894=3.60342\ s

The second ball should be thrown 3.60342 seconds later

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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel
BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

t=\frac{6.22}{2}=3.11 s

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

Solving the  formula, we find

s=\frac{1}{2}(9.8)(3.11)^2=47.4 m

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3 years ago
What are all the<br> invertebrates with a<br> large foot
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Answer:

Explanation:

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3 years ago
A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
Drupady [299]
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
          250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
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3 years ago
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nadya68 [22]

Answer:

attached below

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5 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an
Juliette [100K]

Answer:W=1.93\times 10^9 J      

Explanation:

Given

Force F=1.6\times 10^{6} N

one at an angle of 13^{\circ} East of North and another at 13^{\circ} West of North

Net Force is in North Direction

F_{net}=2F\cos 13

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces  

W=2F\cdot \cos 30\cdot s

here s=0.7 km

W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700

W=1.93\times 10^9 J                  

3 0
3 years ago
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