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Flauer [41]
3 years ago
11

dado los vectores a=3.5 b=4 c=6.5 , si se sabe que a+b+c=0 (resultante) cual es el angulo formado por A y B?

Physics
1 answer:
lawyer [7]3 years ago
6 0

cos(A) = (b2 + c2 − a2) / 2bc = (16+6.5^2-3.5^2)/(2*4*6.5) = 0.88461538  so A = cos−1 (0.88461538) = 27.79 Deg cos(B) = (c2 + a2 − b2)/2ca = (3.5^2+6.5^2-16)/(2*3.5*6.5) = 0.84615385  so B = cos−1 (0.84615385) = 32.2 Deg

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Giving lots of points
bekas [8.4K]

Answer:

wind turbine

Explanation:

5 0
3 years ago
Read 2 more answers
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l
MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

F = -kx

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

F=mg

Inserting this into Hooke's Law

mg=-kx

computing it for x,

-x=mg/k

This is the model which will tell the length of the spring against change in the mass located in the pan.

3 0
3 years ago
How does the input distance of a third-class lever compare to the output distance​
Alexandra [31]

Answer:

A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force

4 0
2 years ago
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