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A spring has a spring constant of 1350 N/m. You place the spring vertically with one end on the floor. You then drop a 1.3 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will be compressed. Express your answer with the appropriate mks units.

Answers

Answer 1

Answer:

Answer:

0.123 m.

Explanation:

From Hook's law,

The potential energy of the book = the energy stored in the spring.

mgh = 1/2ke².................. Equation 1

Where m= mass of the book, g = acceleration due to gravity, h = height, k = spring constant of the spring, e = distance of compression.

make e the subject of the equation

e = √(2mgh/k).................. Equation 2

Given: m = 1.3 kg, h = 0.8 m, k = 1350 N/m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×1.3×0.8×9.8/1350)

e = √(20.384/1350)

e = √(0.0151)

e = 0.123 m.

Answer 2

Answer:

0.015m (downwards)

Explanation:

When the book is dropped on the top of the spring at that height, the potential energy ( $E_(P)$ ) of the book is converted to elastic energy ( $E_(E)$ ) on the spring thereby causing a compression on the spring. i.e

$E_(P)$ = $E_(E)$

But;

The potential energy $E_(P)$ of the mass (book), is the product of the mass(m) of the book, the height(h) from which it was dropped and the acceleration due to gravity (g). i.e

$E_(P)$ = - m x g x h [the -ve sign shows a decrease in height as the mass (book) drops]

Also;

The elastic energy ( $E_(E)$ ) of compression of the spring is given by

$E_(E)$ = $(1)/(2)$ x k x c

Where;

c = compression length of the spring

k = the spring's constant

Substitute these values of $E_(P)$ and E into equation (i) as follows;

- m x g x h = $(1)/(2)$ x k x c ----------------(ii)

From the question;

m = 1.3kg

h = 0.8m

Take g = 10m/s²

k = 1350N/m

Substitute these values into equation (ii) as follows;

- 1.3 x 10 x 0.8 = $(1)/(2)$ x 1350 x c

- 10.4 = 675c

Solve for c;

c = - 0.015 m [The negative sign shows that the spring actually compresses]

Therefore, the maximum distance the spring will be compressed is 0.015m (downwards of course).

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A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.

Answers

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

$\Sigma F = F - m\cdot g = m\cdot a$ (1)

Where:

$F$ - Buoyant force, measured in newtons.

$m$ - Mass of the plastic ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration, measured in meters per square second.

If we know that $F = 5\,N$ , $m = 0.408\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then the net acceleration of the plastic ball is:

$a = (F)/(m) - g$

$a= 2.448\,(m)/(s^(2))$

The acceleration is 2.448 meters per square second and is vertically upward.

A physics cart has a projectile launcher mounted on top. While traveling on a straight track at 0.500 m/s, a projectile is fired. It lands back in the same place on top of the launcher after the cart has moved a distance of 2.30 m. In the frame of reference of the cart, (a) at what angle was the projectile fired and (b) what was the initial velocity of the projectile? (c) What is the shape of the projectile as seen by an observer on the cart? A physics student is watching the demonstration from a classroom seat. According to the student, (d) what is the shape of the projectile’s path, and (e) what is its initial velocity?

Answers

Answer:

(a) $90^(\circ)$

(b) Initial velocity of the projectile is 22.54 m/s

(d) Parabolic

(e) The initial velocity is 23.04 m/s

Solution:

As per the question:

Velocity of the cart, v = 0.500 m/s

Distance moved by the cart, d = 2.30 m

Now,

(a) The projectile must be fired at an angle of $90^(\circ)$ so that it mounts on the top of the cart moving with constant velocity.

(b) Now, for initial velocity, u':

Time of flight is given by;

$T = (D)/(v)$ (1)

where

T = Flight time

D = Distance covered

(b) The component of velocity w.r.t an observer:

Horizontal component, $v_(x) = u'cos\theta$

Vertical component, $v_(y) = u'sin\theta - gT$

Also, the vertical component of velocity at maximum height is zero, $v_(y) = 0$

Therefore, $T = (u')/(g)$

Total flight time, $(2u')/(g)$ (2)

Now, from eqn (1) and (2):

$u' = (gD)/(2v)$

$u' = (9.8* 2.30)/(2* 0.500) = 22.54 m/s$

(d) The shape of the path of the projectile seen by the physics student outside the reference frame of the cart is parabolic

(e) The initial velocity is given by:

u = u' + v = 22.54 + 0.5 = 23.04 m/s

Beverage can is thrown upward and then falls back down to the floor. As usual, a y axis extends upward (positive direction). Which of the following best describes the acceleration of the can during its free flight?a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
b) +9.8 m/s^2 throughout
c) -9.8 m/s^2 throughout
d) zero throughout
e) +9.8 m/s^2, then momentarily zero, then -9.8 m/s^2

Answers

a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s2. How far will the car travel in 12 seconds?

Answers

Same formula as the last question. x = vt + (1/2)at^2. In this case, v = 0, t = 12, and a = 2.0. Plug in the values and solve for x (which is change in position)x = (0)(12) + (1/2)(2.0)(12^2)x = (1/2)(2.0)(144)x = (1)(144)x = 144So the car will travel 144 meters in 12 seconds.

Final answer:

The car, accelerating at a constant rate of 2.0 m/s2 from rest, will travel a distance of 144 meters in 12 seconds.

Explanation:

The question pertains to the concept of motion in physics, specifically how distances travelled are influenced by an object's acceleration. The car is accelerating at a constant rate of 2.0 m/s2 from rest. It means that the initial velocity of the car is 0. We can use the formula of motion, s = ut + 0.5at2, where u is the initial velocity, a is the acceleration and t is the time.

In this case, u = 0 (as the car starts from rest), a = 2.0 m/s2 (constant acceleration) and t = 12 seconds. Substituting these values into the formula, we get:

s = 0*12 + 0.5*2*122

Therefore, the car will travel 144 meters in 12 seconds assuming it accelerates at a constant rate.

Learn more about Constant Acceleration here:

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A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but the plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?

Answers

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

$(V_1)/(T_1)=(V_2)/(T_2)\n\Rightarrow {V_2}=(V_1)/(T_1)* T_2\n\Rightarrow {V_2}=(12)/(298.15)* 373.15\n\Rightarrow {V_2}=15.01 L$

∴ Final volume at 100°C is 15.01 Liters.

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Answers

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

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