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15 days ago

A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d

egrees south of that direction, and to continue to drive 50 miles deviating by 30 degrees each time until he returns to his original position. How far will he drive, and how many vectors must he sum to calculate his displacement

Physics

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The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu

Keith_Richards [3271]

Answer:

The greatest force is F = mg.

Explanation:

Applying Newton’s second law:

Σ F = m a

(Vectors are in bold, and we analyze components along x and y axes.)

Along the x-axis:

Fₓ = maₓ

Along the y-axis:

F_y - W = m a_y $a_{y}$

In this scenario, where the bird remains level, the wing force acts at an angle to the x-axis. The vertical component creates lift. Using trigonometry, the components are:

Cos θ = Fₓ / F ⇒ Fₓ = F cos θ

Sin θ = F_y / F ⇒ F_y = F sin θ

Substituting into the vertical force equation:

F sin θ - w = m a_y

Since the bird is hovering at the same height, vertical acceleration is zero (a_y = 0):

F sin θ = w = mg

The maximum lift occurs when sin θ = 1, thus:

F = mg

3 0

4 months ago

The figure above represents a stick of uniform density that is attached to a pivot at the right end and has equally spaced marks

Sav [3153]

Answer:

After a duration of 2.0 seconds, the angular momentum of the system is $L= 2(4A+3B+2C+D)x$ .

Explanation:

Let’s denote the forces acting on the rod as A, B, C, and D, and the distance between them as $x$ .

The torque produced by force A can be expressed as

$\tau_a = 4Ax$ ,

for force B the torque is

$\tau_b = 3Bx$ ,

the torque for force C is

$\tau_c = 2Cx$ ,

and the torque due to force D is

$\tau_d = Dx$ .

This leads to a total torque on the stick as

$\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d$

$\tau_{tot} =4Ax+3Bx+2Cx+Dx$

$\tau_{tot} =x(4A+3B+2C+D)$

Subsequently, this torque results in an angular acceleration $\alpha$ in accordance with the formula

$I \alpha = \tau_{tot}$

where $I$ represents the moment of inertia of the stick, which has a value of

$I = \dfrac{1}{3} m(4x)^2$ .

Thus, the angular acceleration calculates to

$\alpha = \dfrac{\tau_{tot} }{I}$

$\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }$

$\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}$ .

Now, the angular momentum $L$ of the rod is given by

$L = I\omega$ ,

where $\omega$ is considered the angular velocity.

Given $\omega = \alpha t$ , we conclude that

$L = \dfrac{1}{3}m (4x)^2 *\dfrac{3(4A+3B+2C+D)}{16mx }* t$

$L= (4A+3B+2C+D)x t$

Thus, $t = 2.0s$ , the angular momentum is

$\boxed{ L= 2(4A+3B+2C+D)x. }$

5 0

4 months ago

Four particles with masses 2 kg, 5 kg, 2 kg, and 2 kg are connected by rigid rods of negligible mass as shown. assume the system

Maru [3345]

The answer is 10pi. I believe this will be helpful.

8 0

2 months ago

Six pendulums of mass m and length L as shown are released from rest at the same angle (theta) from vertical. Rank the pendulums

Sav [3153]

Answer: 1m, 1m, 2m, 2m, 4m, 4m.

It’s important to remember that the masses attached do not influence the number of oscillations.

Explanation:

To determine the number of oscillations (complete cycles), we can apply the formula n = t / T ……equation 1

The variables that impact the period of a simple pendulum are solely its length and gravitational acceleration. The period remains unaffected by factors such as mass.

period (T)= 2 x π x √(L/g) ….equation 2

where π = 3.142, L= rope length, and g = 9.8 m/s (gravitational acceleration)

According to the question, the time (t) is 60 seconds.

By merging equations 1 and 2, we obtain

number of oscillations = time / (2 x π x √(L/g))

Case 1: for L = 4m

number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))

= 14.9 = 14 complete cycles (the problem specifies complete cycles)

Case 2: for L = 2m

number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))

= 21.4 = 21 complete cycles

Case 3: where L = 4m, results in the same as case 1, yielding 14 complete cycles

Case 4: where L = 2m, mirrors the outcome in case 2, producing 21 complete cycles

Case 5: in the instance of L = 1m

number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))

= 30.1 = 30 complete cycles

Case 6: when L = 1m, which repeats case 5, also gives 30 complete cycles

From these findings, the order of the pendulums from the highest to lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.

Remember, the number of oscillations is independent of their respective masses.

3 0

3 months ago

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