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Vanyuwa [196]
3 years ago
15

1.) Starting from rest, a car accelerates at 6.52 m/s^2 for 3.80 s. Determine the distance traveled by the car during this time.

Physics
1 answer:
inysia [295]3 years ago
3 0

Answer:

47.1 m

Explanation:

First, write the information you're given.

v₀ = 0 m/s

a = 6.52 m/s²

t = 3.80 s

Identify the variable you're looking for:

Find: Δx

Next, find the kinematic equation you need to solve.  One trick is to identify which variable you <em>don't</em> have, then find the equation that doesn't use that variable.  In this case, we don't have the final velocity v, so the equation we should use is:

Δx = v₀ t + ½ at²

Plug in the values and solve:

Δx = (0 m/s) (3.80 s) + ½ (6.52 m/s²) (3.80 s)²

Δx = 47.1 m

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What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
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Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

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C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N
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Answer:

Spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.

Explanation:

Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in <u>elastic region</u>, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.

Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.

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A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
dalvyx [7]

Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

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The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

(C) The direction of the force makes will be

tanθ = 0.172/0.68

θ = 14.19° to the x axis

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