A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?
Answers
Answer:
1) the required horizontal force F is 1095.6 N
2) W = 0 J { work done by rope will be 0 since tension perpendicular }
3) work is done by the worker is 1029.4 J
Explanation:
Given that;
mass of bag m = 125 kg
length of rope [tex]l[/tex] = 3.3 m
displacement of bag d = 2.2 m
1) What horizontal force is necessary to hold the bag in the new position?
from the figure below; ( triangle )
SOH CAH TOA
sin = opp / hyp
sin[tex]\theta[/tex] = d / [tex]l[/tex]
sin[tex]\theta[/tex] = 2.2/ 3.3
sin[tex]\theta[/tex] = 0.6666
[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )
[tex]\theta[/tex] = 41.81°
Now, tension in the string is resolved into components as illustrated in the image below;
Tsin[tex]\theta[/tex] = F
Tcos[tex]\theta[/tex] = mg
so
Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg
sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg
we know that; tangent = sine/cosine
so
tan[tex]\theta[/tex] = F / mg
F = mg tan[tex]\theta[/tex]
we substitute
Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )
F = 1225 × 0.8944
F = 1095.6 N
Therefore, the required horizontal force F is 1095.6 N
2) As the bag is moved to this position, how much work is done by the rope?
Tension in the rope and displacement of mass are perpendicular,
so, work done will be;
W = Tdcos90°
W = Td × 0
W = 0 J { work done by rope will be 0 since tension perpendicular }
3) As the bag is moved to this position, how much work is done by the worker
from the diagram in the image below;
SOH CAH TOA
cos = adj / hyp
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex]
we substitute
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex] = 1 - h/[tex]l[/tex]
cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]
h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]
h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
now, work done by the worker against gravity will be;
W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
we substitute
W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )
W = 4042.5 × ( 1 - 0.745359 )
W = 4042.5 × 0.254641
W = 1029.4 J
Therefore, work is done by the worker is 1029.4 J
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Explanation:
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g be gravitational acceleration 9.81ms^-2
They share same T and a
by force diagram the net force on m:
T-2.1g=2.1a
net force on M:
6.3g-T=6.3a
solve:
(T-2.1g)/(6.3g-T)=2.1a/6.3a
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Explanation:
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Answers
Answer:
v = 3.74 m/s
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Given that,
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Hi there!
[tex]\large\boxed{\text{C. 17.7m/s}}[/tex]
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vf² = vi² + 2(ad)
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Answers
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=
Explanation: I REALLY HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]
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Given parameters:
Mass of the object: m = 20 kg.
Acceleration of the object: a = 10.0 m/s^2.
Hence, according to Newton's 2nd law of motion:
the magnitude of requires force = mass ×acceleration
= 20 × 10 Newton
= 200 Newton.
Hence, the magnitude of requires force is 200 Newton.
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https://brainly.com/question/13191643
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The period is less, but not halved.
The period is more, but not doubled.
The period is halved.
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Answers
Answer:
The period is more, but not doubled.
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Answers
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While sitting in a boat, a fisherman observes that 2 complete waves pass by his position every 4 seconds. What is the period of these waves?A)2s B)8s C) 0.5 s D) 4s
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Answer:
2
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Because 2 waves 4 secs means 1 in 2s
The period of these waves is 2s. and The right option is A)2s.
The period of a wave is the time taken by a wave to complete one cycle.
The formula of period from the question is given below.
⇒ Formula:
T = t/2.................... Equation 1⇒ Where:
T = Period of the wavet = time taken for two complete oscillationFrom the question,
⇒ Given:
t = 4 seconds.⇒ Substitute these value above into equation 1
T = 4/2T = 2 seconds.Hence, The period of these waves is 2s.
The right option is A)2s
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Explanation:
Projectile motion, Bernoulli's principle, and Magnus Effect.
Sure I would be happy to discuss projectile motion!
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Answer:
Weight
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A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.80 T in 1.10 s. What is the resulting induced current if the loop has a resistance of 2.20
Answers
Answer:
The magnitude of the induced current is 4.73 x 10⁻³ A.
Explanation:
Given;
number of turns, N = 1
cross sectional area of the loop, A = 8.8 cm² = 8.8 x 10⁻⁴ m²
change in magnetic field strength, ΔB = 1.8 T - 0.5 T = 1.3 T
change in time, Δt = 1.10 s
resistance of the loop, R = 2.2 ohm
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA \Delta B}{\Delta t} \\\\emf = \frac{1 \times 8.8\times 10^{-4} \times 1.3}{1.10} \\\\emf = 1.04 \times 10^{-3} \ V[/tex]
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Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Answers
Answer:
Please see below as the answer is self-explanatory.
Explanation:
We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).Both movements are independent each other, due to they are perpendicular.In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:[tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]
Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:[tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]
In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]
In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:[tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]
Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:[tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]
Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.An object is spun in a circle of radius 1.5m with a frequency of 6.0Hz. what is it’s velocity ?
Answers
Answer:
56.55 m/s
Explanation:
Frequency=1/T
1/6 = 0.1666666
V=(2*pi*1.5)/0.166666
An object is moving with constant non-zero velocity. Which of thw following statements about it must be true
Answers
Answer:
The net force on the object is zero.
Explanation:
An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".