A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.8 m into the house.
1 answer:
The work done by a force is equal to:
where
W is the work
F is the force applied
d is the distance through which the force is applied.
In our problem, the force needed to lift the dog is equal to the weight of the dog:
F=24 N
and the dog is moved through a distance of
d=4.8 m
therefore, the work done by the child to lift the dog is
where
W is the work
F is the force applied
d is the distance through which the force is applied.
In our problem, the force needed to lift the dog is equal to the weight of the dog:
F=24 N
and the dog is moved through a distance of
d=4.8 m
therefore, the work done by the child to lift the dog is
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Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.
Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
+
+
=
+
Ф
60 ·8 + 0 = (60 + 80)Ф
480 = 140Ф................. (I)
y-axis component form (+y north);
+
+
=
+
Ф
0 + 80.9 = (60 + 80)Ф
720=
140Ф
140Vf=Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.
Answer:
The separation between the first two minima on either side is 0.63 degrees.
Explanation:
A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:
with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:
for the first minimum
solving for θ1:
for the second minimum:
So, the angular separation between them is the rest: