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2 years ago

The upward force acting on the wing of an airplane ir flight is called?

Physics

2 answers:

Elza [17]2 years ago

7 0

Answer:

Lift is the upward force on the wing acting perpendicular to the relative wind and perpendicular to the aircraft's lateral axis. Lift is required to counteract the aircraft's weight.

Explanation:

sp2606 [1]2 years ago

5 0

Answer: The upward force acting on the wing of an airplane in flight is called <u>Lift</u>

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A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

$W_f = F_fs = 15*0.5 = 7.5 J$

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

$E_p = mgh = 0.1*10*1.5 = 1.5 J$

Therefore, as elastic energy is converted to potential energy and work of friction:

$E_e = W_f + E_p$

$kx^2/2 = 7.5 + 1.5 = 9 J$

$k = 9*2/x^2 = 18/0.05^2 = 7200 N/m$

6 0

3 years ago

Number of valence electrons for calcium?

zimovet [89]

Calcium has 2 valence electrons

7 0

3 years ago

Read 2 more answers

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

A train increases its speed steadily from 10 m/s to 20 m/s in

rewona [7]

Answer:

15m/s

Explanation:

add the two speeds and divide by 2

10+20=30

30/2=15

3 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$