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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1

luda_lava [24]

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

4 0

2 years ago

Kinetic Energy - What does it depend on?

Greeley [361]

Answer:

faster; more kinetic energy

Explanation:

7 0

2 years ago

Two tractors pull against a 1000 kg log. If the angle of the tractors' chains in relation to each other is 18.0° (each 9.0° from

Morgarella [4.7K]

Pie ?????????????? π pizza ??? wut

7 0

3 years ago

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

3 years ago

A 1.90-kg mass vibrating up and down on the end of a vertical spring has a maximum speed of 2.30 m/s. What is the total potentia

Pepsi [2]

Answer:

The answer to the question is;

The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.

Explanation:

To answer the question, we note that the maximum speed is 2.30 m/s and the mass is 1.90 kg

Therefore the maximum kinetic energy of motion is given by

Kinetic Energy, KE = $\frac{1}{2} mv^{2}$

Where,

m = Attached vibrating mass = 1.90 kg

v = velocity of the string = 2.3 m/s

Therefore Kinetic Energy, KE = $\frac{1}{2}$ ×1.9×2.3² = 5.0255 J

From the law of conservation of energy, we have the kinetic energy, during the cause of the vibration is converted to potential energy when the mass is at either endpoint of its motion

Therefore Potential Energy PE at end point = Kinetic Energy, KE at the middle of the motion

That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.

Total PE = Maximum KE = 5.0255 J.

6 0

2 years ago