PHYSICS COLLEGE

A 50 kg child on a skateboard experiences a 75-N force as shown.What is the expected acceleration of the child?

F = 75 N
7
.

A. 0.67 m/s2
B. 1.50 m/s2
C. 6.70 m/s2
D. 25.0 m/s2

Answers

Answer 1

Answer:

Answer:

1.50 m/s²

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = (f)/(m) \n$

f is the force

m is the mass

From the question we have

$a = (75)/(50) = (3)/(2) \n$

We have the final answer as

1.50 m/s²

Hope this helps you

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A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. If the spring constant is 1050 N/m and the mass of the rocket is 50 g, how high will the rocket go? You may neglect the effects of air resistance.
Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).
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A 1.5v battery stores 4.5KJ of energy. How long can it light a flashlight bulb that draws 0.60A

A 0.26-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released?

Answers

Complete Question

The

Answer:

$E_r = 3.058 \ J$

$E_b = -11.466 \ J$

$\Delta E_n = -14.524 \ J$

Explanation:

From the question we are told that

The mass of the stone is $m_s = 0.26 \ kg$

The height above the top of the water is $h = 1.2 \ m$

The depth of the well is $d = 4.5 \ m$

The gravitational potential of the stone before it was released is

$E_r = mgh$

substituting values

$E_r = 0.26 * 9.8 * 1.2$

$E_r = 3.058 \ J$

The gravitation potential of the stone when it reaches the bottom of the well is

$E_b = mg(- d)$

The negative shows that the potential energy of the stone as compared to the earth is reducing

substituting values

$E_b = 0.26 * 9.8 *(- 4.5)$

$E_b = -11.466 \ J$

The change in the systems gravitational potential is

$\Delta E_n = E_b - E_r$

substituting values

$\Delta E_n = -11.466 - 3.058$

$\Delta E_n = -14.524 \ J$

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?

Answers

The linear speed of the ball for the circular motion is determined as 12 m/s.

The given parameters;

mass of each ball, m = 450 g = 0.45 kg
length of the rod, L = 0.2 m
radius of the rod, r = 0.1 m
angular speed of the ball, ω = 120 rad/s

The linear speed of the ball is calculated as follows;

v = ωr

where;

ω is the angular speed of the ball
r is the radius of circular motion of the ball

The linear speed of the ball is calculated as follows;

v = ωr

v = 120 x 0.1

v = 12 m/s

Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.

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Answer:

The speed of ball is 12 $(m)/(s)$

Explanation:

Given:

Mass of ball $m = 0.45$ kg

Radius of rotation $r = 0.1$ m

Angular speed $\omega = 120 (rad)/(s)$

Here barbell spins around a pivot at its center and barbell consists of two small balls,

From the formula of speed in terms of angular speed,

$v = r \omega$

Where $v =$ speed of ball

$v = 120 * 0.1$

$v = 12 (m)/(s)$

Therefore, the speed of ball is 12 $(m)/(s)$

(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).

Answers

Answer:

k = 15.62 MN/m

Explanation:

Given:-

- The viscous damping constant, c = 1.8 KNs/m

- The floor oscillation magnitude, Yo = 3 mm

- The frequency of floor oscillation, f = 18 Hz.

- The combined weight of the grinding machine and the wheel, W = 4200 N

- Two springs of identical stiffness k are attached in parallel arrangement.

Constraints:-

- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )

Find:-

What is the minimum required stiffness of each of the two springs as per the constraints given.

Solution:-

- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

$y ( t ) = Y_o*sin ( w*t )$

Where,

Yo : The amplitude of excitation = 3 mm

w : The excited frequency = 2*π*f = 2*π*18 = 36π

- The harmonic excitation of the floor takes the form:

$y ( t ) = 3*sin ( 36\pi *t )$

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

$m*(d^2x)/(dt^2) + c*(dx)/(dt) + k_e_q*x = k_e_q*y(t) + c*(dy)/(dt)\n\n(m)/(k_e_q)*(d^2x)/(dt^2) + (c)/(k_e_q)*(dx)/(dt) + x = y(t) + (c)/(k_e_q)*(dy)/(dt)$

Where,

m: The combined mass of the rigid body ( wheel + grinding wheel body) c : The viscous damping coefficient

k_eq: The equivalent spring stiffness of the system ( parallel )

x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

$w_n = \sqrt{(k_e_q)/(m) } , \n\np = (c)/(2√(k_e_q*m) ) = (1800)/(2√(k_e_q*428.135) ) = (43.49628)/(√(k_e_q) )$

Where,

w_n: The natural frequency

p = ζ = damping ratio = c / cc , damping constant/critical constant

- The Equation of motion becomes:

$(1)/(w^2_n)*(d^2x)/(dt^2) + (2*p)/(w_n)*(dx)/(dt) + x = y(t) + (2*p)/(w_n)*(dy)/(dt)$

- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

$X_s_s = X_o*sin ( wt + \alpha )$

Where,

X_o : The amplitude of the steady-state vibration.

α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

$X_o = Y_o*\sqrt{(1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2) }$

Where,

r = Frequency ratio = $(w)/(w_n) = \frac{36*\pi }{\sqrt{(k_e_q*g)/(W) } } = \frac{36*\pi }{\sqrt{(k_e_q)/(428.135) } } = (36*\pi*√(428.135) )/(√(k_e_q) )$

- We will use the one of the constraints given to limit the amplitude of steady state oscillation ( Xo ≤ 10 mm ):

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):

$((X_o )/(Y_o))^2 \geq (1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2)\n\n((X_o )/(Y_o))^2 \geq (1+ ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)/(( 1 - ((36*\pi*√(428.135) )/(√(k_e_q) ))^2)^2 + ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)\n\n$

$((X_o )/(Y_o))^2 \geq ( 1 + (41442858448.85813)/(k_e_q^2 ))/([ 1 - ((5476277.91201 )/(k_e_q) )]^2 + (41442858448.85813)/(k_e_q^2 ) )}\n\n((X_o )/(Y_o))^2 \geq ( (k_e_q^2 + 41442858448.85813)/(k^2_e_q ))/([ ((k_e_q - 5476277.91201)^2 )/(k_e_q^2) ] + (41442858448.85813)/(k_e_q^2 ) )}\n$

$((X_o )/(Y_o))^2 \geq ( k_e_q^2 + 41442858448.85813)/( (k_e_q - 5476277.91201)^2 +41442858448.85813 )}\n\n((10 )/(3))^2 \geq ( k_e_q^2 + 41442858448.85813)/( k^2_e_q -10952555.82402*k_e_q +3.00311*10^1^3 )}\n\n\n10.11111*k^2_e_q -121695064.71133*k_e_q +3.33637*10^1^4 \geq 0$

- Solve the inequality ( quadratic ):

$k1_e_q \geq 7811740.790197058 (N)/(m) \n\nk2_e_q \leq 4224034.972855095 (N)/(m)$

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

$k_e_q = (k^2)/(2k) = (k)/(2)$

- Therefore,

$k1 \geq 7811740.790197058*2 = 15.62 (MN)/(m) \n\nk2 \leq 4224034.972855095*2 = 8.448 (MN)/(m)$

- The minimum stiffness of spring is minimum of the two values:

k = 15.62 MN/m

Mudflows composed of soil, volcanic debris, and water can occur as the result of an explosive volcanic eruption. What are these mudflows called?

Answers

These mudflows are called Lahar.

A quickly moving mixture of rock debris and water that begins on a volcano's slopes is referred to as a lahar in Indonesian. Other names for lahars are volcanic mudflows and debris floods. The size, pace, and volume of material transported by a moving lahar can constantly alter as it rushes downstream. It resembles a swirling slurry of wet concrete.

The melting of snow and ice as well as the ingestion of river or lake water by the moving slurry may both add to its water consumption. A lahar's starting flow may be quite tiny, but as it entrains and integrates everything in its path, including rocks, dirt, vegetation, even structures like houses and bridges, it may increase in volume.

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Which of the following is a description of the Remote Associates Test (RAT)?

Answers

Answer:

The description is outlined throughout the clarification section following, and according to the given word.

Explanation:

Throughout the 1960s, Sarnoff Mednick created the RAT as a tool used for testing imaginative convergent thought. Through each RAT test query lists a set of terms, which demands that we have a single additional term that will tie any of the others around.
Those other words may also be related in something like a variety of ways, such as through creating a compound word or perhaps a semantic connexon.

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?