The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
Answers
We have that for the Question, it can be said that With respect to this axis, the magnitude of the torque due to the weight and ,the thrust is
TW=19740N-mTT=130387.39N-mFrom the question we are told
The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
a)
Generally the equation for the Torque due to weight is mathematically given as
[tex]TW=Engine weight*2.50*sin32\\\\TW=14900*2.50*sin32[/tex]
TW=19740N-m
b)
Generally the equation for the Torque due to thrust is mathematically given as
[tex]TT=Engine thrust*2.50*cos32\\\\TT=61500*2.50*cos32\\\\[/tex]
TT=130387.39N-m
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Answer:
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When converted to a household measurement, 9 kilograms is approximately equal to a
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Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
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A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s . Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.
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the polar bear will travel 302.5m
a block starts from rest and begins sliding down an incline. The block reaches a speed of 12 meters per second as it slides a distance of 50 meters. calculate the blocks rate of acceleration
Answers
Vf² = V₁² + 2ad
(12 m/s) ² = (0 m/s) ² + 2 (a) (50m)
a = 1.44 m/s ²
Answer:
[tex]\boxed {\boxed {\sf 1.44 \ m/s^2}}[/tex]
Explanation:
We are asked to find the block's rate of acceleration.
We are given the final velocity, initial velocity, and distance. Therefore, we will use the following kinematic equation:
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The block starts at rest or 0 meters per second. The block reaches a final velocity of 12 meters per second. The distance traveled is 50 meters.
[tex]v_i[/tex]= 0 m/s [tex]v_f[/tex]= 12 m/s [tex]d[/tex]= 50 mSubstitute the values into the formula.
[tex](12 \ m/s)^2 = (0 m/s)^2 + 2a (50 \ m)[/tex]
Solve the exponents.
(12 m/s)² = 12 m/s * 12 m/s = 144 m²/s²(0 m/s)² = 0 m/s * 0 m/s = 0 m²/s²[tex]144 \ m^2/s^2= 0 m^2/s^2 + 2a (50 \ m)[/tex]
The 0 can be subtracted from both sides, or simply canceled.
[tex]144 \ m^2/s^2 = 2a (50 \ m)[/tex]
Multiply 2 and 50 meters.
[tex]144 \ m^2/s^2 = (2 *50 \ m)* a[/tex]
[tex]144 \ m^2/s^2 = (100 \ m)* a[/tex]
We are solving for a, so we must isolate the variable. It is being multiplied by 100 meters. The inverse of multiplication is division, so we divide both sides by 100 meters.
[tex]\frac {144 \ m^2/s^2}{100 \ m}= \frac{(100 \ m)*a}{100 \ m}[/tex]
[tex]\frac {144 \ m^2/s^2}{100 \ m}=a[/tex]
[tex]1.44 \ m/s^2 =a[/tex]
The block's rate of acceleration is 1.44 meters per second squared.
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the snowball before it is released.
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b) 19.6J
c) 5J
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Answer: B 19.6J
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PE is the potential energy
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h=Height=5 metres
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Answer:
low pressure
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The S.I unit of solid angle is steradian, or square radian (symbol: sr).
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12414253
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Answer:
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a) What is the friction force acting on the mass?
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Answer:
Below
Explanation:
a) To find the force of friction acting on the mass we can use this formula :
friction = coefficientoffriction x normal force
First we need to find the normal force of this object :
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Now we can find the force of friction acting on the object :
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Answer:
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Explanation:
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Answer:
Explanation:
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Formula
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m = 100 kg
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Answer:
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Explanation:
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Answer:
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Answer:
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Answer:
4. pressure is defined as force per unit area
Pressure is defined by force per unit area.
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Answer:
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Explanation:
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
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Answers
We have that for the Question "Light of 650 nm wavelength illuminates two slits that are 0.20 mm apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?" it can be said that the distance to the screen
d=1.168m
From the question we are told
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
Generally the equation for the distance is mathematically given as
[tex]d=\frac{0.33*10^{-2}*0.23*10^{-3}}{650*10^{-9}}\\\\d=1.160m[/tex]
d=1.168m
Therefore
The distance to the screen
d=1.168m
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A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introduction of 430 J of heat caused the system to expand, doing 238 J of work against a constant pressure of 101 kPa. What is the value of for this process
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Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
what is 4 differences between saturated unsaturated and supersaturated solutions
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Answer:
Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.
Bill is sitting on a tree limb that is 4 meters above the ground with a water balloon. He wants to drop the water balloon so that it falls on his friend Tim who is approaching on a bike at a constant speed of 8 meters per second. How far away should Tim be from a point directly under bill when bill drops the balloon?
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Answer:
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Answer:
Do you have. a picture or full page for you're question its saying following parts
A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonconducting 2.50 cm wire. A uniform electric field of magnitude 1.85×108N/C is directed parallel to the wire, as shown in the figure. (Figure 1)
Part A
Find the tension in the wire.
Express your answer with the appropriate units.
Part B
What would the tension be if both charges were negative?
Express your answer with the appropriate units.
Answers
(a) The tension on the wire when the two charges have opposite signs is 383.5 N.
(b) The tension on the wire if both charges were negative is 3.640.25 N.
The given parameters;
first charge, q₁ = 8.75 μC second charge, q₂ = -6.5 μC electric field, E = 1.85 x 10⁸ N/Cdistance between the two charges, r = 2.5 cm(a)
The attractive force between the charges is calculated as follows;
[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N[/tex]
The force on the negative charge due to the electric field is calculated as follows;
[tex]F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N[/tex]
The tension on the wire is the resultant of the two forces and it is calculated as follows;
[tex]T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N[/tex]
(b) when the two charges are negative
The repulsive force between the two charges is calculated as follows;
[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N[/tex]
The force on the first negative charge due to the electric field is calculated as follows;
[tex]F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N[/tex]
The force on the second negative charge due to the electric field is calculated as follows;
[tex]F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N[/tex]
The tension on the wire is the resultant of the three forces and it is calculated as follows;
[tex]T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N[/tex]
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giúp em toàn bộ nhé ............
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Explanation:
nnnnkgcbchg of bbjbjk oh jjzjjzkedkkdkfjjsjdsjfkjfdkkdowkronqfojofj see j FC