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emmasim [6.3K]
3 years ago
10

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal

l be in the air before it lands?
Round your answer to the nearest tenth.
Physics
1 answer:
alukav5142 [94]3 years ago
5 0
Just consider the vertical component
d_y=v_{0,y}t+\frac{1}{2}at^2
d_y=(13.9\sin 25\°)t+\frac{1}{2}(-9.8~m/s^2)t^2 \\  \\ d_y=5.87t-4.9t^2

Set d_y equal to 0 to find the times the ball is on the ground
0=5.87t-4.9t^2 \\  \\ 0=t(5.87-4.9t) \\  \\ t=0 \\  \\ t=1.2~s

The ball was in the air for 1.2 seconds
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What unit is used for speed?<br> 1. m/s<br> 2. S<br> 3. m/s2<br> 4. m
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What is universal laws of gravitation ​
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3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

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2 years ago
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