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4 years ago

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On takeoff, the combined action of the engines and wings of an airplane exerts a(n) 7934 N force on the plane, directed upward a

t an angle of 54.2 ◦ above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. The acceleration of gravity is 9.8 m/s 2 . What is the weight of the plane? Answer in units of N.

1 answer:

mrs_skeptik [129]4 years ago

7 0

Answer:

$W=6435N$

Explanation:

From the exercise we know that the force on the plane is 7934N upward at an angle of 54.2º.

If analyze the forces in the vertical direction, the acceleration is zero because the plane is moving with constant velocity

∑ $F_{y}=7934sin(54.2)-W=0$

$W=7934sin(54.2)N=6435N$

So, the <u><em>weight of the plane</em></u> is 6435N

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Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

Step2247 [10]

Answer:

$\Delta p = 90.7 kPa$

Explanation:

specific gravity of oil is $= \frac{\rho_{oil}}{\rho_w}$

$\rho_{oil} = 0.85*1000 = 850 kg/m3$

we know that

change in pressure for oil is given as

$\Delta p = \rho gh$

here density and h is for oil

$\Delta p = 850*5 *9.81 = 41,692.5 kPa$

change in pressure for WATER is given as

$\Delta p = \rho gh$

here density is for water and h is for water

$\Delta p = 1000*5 *9.81 = 49,050 kPa$

pressure change due to both is given as

$\Delta p = 41692.3 + 49050 = 90742.5 N/m2$

$\Delta p = 90.7 kPa$

8 0

3 years ago

Read 2 more answers

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

4 0

3 years ago

Data that shows consistent, regular, repetitive form displays a pattern.<br><br> True<br><br> False

anyanavicka [17]

I think the answer is True

6 0

3 years ago

Read 2 more answers

In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the

MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 2.5 kg$ is the mass of the ball

$u_1 = 7.5 m/s$ is the initial velocity of the ball

$m_2 = 70 kg$ is the mass of the man

$u_2 = 0$ is the initial velocity of the man

$v$ is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

$v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s$

So, the man and the ball slides on the ice at 0.259 m/s.

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3 0

3 years ago

Can something have energy without having momentum? explain. can something have momentum without having energy? defend your answe

marta [7]

Momentum is a product mass and velocity. If a certain object posses a kinetic energy, then it should have a momentum since it is moving which has a velocity. However, if the object is at rest and only has potential energy, then it would not have momentum. So, for the first question the answer would be yes, an object can have energy without having any momentum. For the second question, every object whether it is moving or at rest, possess some energy, potential for an object at rest and kinetic for an object that is moving. Thus, the answer would be no, an object having momentum would always have energy.

8 0

3 years ago