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Xelga [282]
2 years ago
12

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

is blowing. If the wind's force on the rope is negligible, what drag force in Newtons does the wind exert on the ball? B)A box is sliding down an incline tilted at a 15° angle above horizontal. The box is initially sliding down the incline at a speed of 1.4 m/s. The coefficient of kinetic friction between the box and the incline is 0.37. How far does the box slide down the incline before coming to rest?C)An object weighing 3.9 N falls from rest subject to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b = 2.5 N ∙ s2/m2. What terminal speed will this object approach?
Physics
1 answer:
Marat540 [252]2 years ago
7 0

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

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Answer:

Fossil fuels release carbon dioxide into the atmosphere. Natural gas is the best fossil fuel in terms of energy output per unit of carbon dioxide emitted. Biomass is renewable because a new crop can be grown after each harvest, and biomass is a low carbon fuel. Biomass has the approximate chemical formula CHO.

Explanation:

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3 years ago
Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h
PolarNik [594]

Answer:

(a) a = - 201.8 m/s²

(b) s = 197.77 m

Explanation:

(a)

The acceleration can be found by using 1st equation of motion:

Vf = Vi + at

a = (Vf - Vi)/t

where,

a = acceleration = ?

Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)

Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s

t = time = 1.4 s

Therefore,

a = (0 m/s - 282.53 m/s)/1.4 s

<u>a = - 201.8 m/s²</u>

<u></u>

(b)

For the distance traveled, we can use 2nd equation of motion:

s = Vi t + (0.5)at²

where,

s = distance traveled = ?

Therefore,

s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²

s = 395.54 m - 197.77 m

<u>s = 197.77 m</u>

6 0
3 years ago
A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration
Pani-rosa [81]

The magnitude of acceleration is (change in speed) / (time for the change).

Change in speed = (speed at the end) - (speed at the beginning) =

                                   (16 m/s)  -  (0)  =  16 m/s .

Time for the change  =  4 s .

Magnitude of acceleration = (16 m/s) / (4 s) = 4 m/s per sec = 4 m/s² .


6 0
3 years ago
Read 2 more answers
During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
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Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th
vladimir1956 [14]

Answer:

h = 22.35 m

Explanation:

given,

initial speed of the rock,u = 0 m/s

length of the window,l = 2.7 m

time taken to cross the window,t = 0.129 s

Speed of the rock when it crosses the window

v = \dfrac{l}{t}

v = \dfrac{2.7}{0.129}

  v = 20.93 m/s

height of the building above the window

using equation of motion

v² = u² + 2 g h

20.93² = 0² + 2 x 9.8 x h

h = 22.35 m

Hence, the height of the building above the top of window is equal to h = 22.35 m

8 0
2 years ago
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