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3 years ago

Should the hypothesis always be correct for a conclusion

1 answer:

4 0

No it shouldn't, a hypothesis doesn't need to be correct but must have an idea for why x variable effects y variable and have good reasoning. In the conclusion you should state if it's correct or not and explain why it's correct/incorrect and what answer you've determined from data.

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The spring has a constant of 29 N/m and the frictional surface is 0.4 m long with a coeﬃcient of friction µ = 1.65. The 7 kg blo

densk [106]

Answer:

The block lands 3 m from the bottom of the cliff.

Explanation:

Hi there!

(atteched find a figure representing the situation of the problem).

To solve this problem let´s use the theorem of conservation of energy.

Initially, the object has elastic (EPE) and gravitational potential energy (PE):

PE = m · g · h

EPE = 1/2 · k · x²

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression of the spring.

At the bottom of the cliff, this total energy, minus some energy that will be dissipated by friction during the 0.4 m displacement over the frictional surface, will be converted into kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the block

v = velocity of the block.

The work done by friction (Wf) is equal to the dissipated energy:

Wf = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = μ · N = μ · m · g

Where:

N = normal force.

g = acceleration due to gravity.

Then, the final kinetic energy can be calculated as follows:

EPE + PE - Wf = KE

EPE = 1/2 · k · x²

EPE = 1/2 · 29 N/m · (0.19 m)²

EPE = 0.52 J

PE = m · g · h

PE = 7 kg · 9.8 m/s² · (2.8 m + 1m)

PE = 260.7 J

Wf = μ · m · g · d

Wf = 1.65 · 7 kg · 9.8 m/s² · 0.4 m

Wf = 45.3 J

Then:

KE = 0.52 J + 260.7 J - 45.3 J

KE = 215.9 J

Then, we can calculate the magnitude of the velocity when the block reaches the ground:

KE = 1/2 · m · v²

215.9 J = 1/2 · 7 kg · v²

v² = 215.9 J · 2 / 7 kg

v = 7.9 m/s

The time it takes the block to reach the ground from the second drop, can be calculated with the following equation:

h = h0 + v0y · t + 1/2 · g · t²

Where:

h = height at time t.

h0 = initial height.

v0y = initial vertical velocity.

g = acceleration due to gravity.

t = time.

When the block reaches the ground its height is zero. Initially, the block does not have vertical velocity, then, v0y = 0. The initial height is 1 m. Considering the upward direction as positive, the acceleration of gravity is negative:

h = h0 + v0y · t + 1/2 · g · t²

0 m = 1 m + 0 · t - 1/2 · 9.8 m/s² · t²

-1 m = -4.9 m/s² · t²

t² = -1 m / -4.9 m/s²

t = 0.45 s

The vertical velocity (vy), when the block reaches the ground can now be calculated:

vy = v0y + g · t

vy = -9.8 m/s² · 0.45 s

vy = -4.4 m/s

And now, we can finally find the horizontal velocity (vx) of the block. The magnitude of the velocity when the block reaches the ground is calcualted as follows:

$v = \sqrt{ vx^{2} + vy^{2} }$

v² = vx² + vy²

v² - vy² = vx²

√(v² - vy²) = vx

vx = √((7.9 m/s)² - (4.4 m/s)²)

vx = 6.6 m/s

Since there is no force accelerating the block in the horizontal direction, the horizontal velocity of the block when it lands is equal to the initial horizontal velocity. Then, we can calculate the horizontal traveled distance:

x = x0 + v · t (x0 = 0 because we consider the edge of the cliff as the origin of the frame of reference).

x = 0 + 6.6 m/s · 0.45 s

x = 3 m

The block lands 3 m from the bottom of the cliff.

4 0

4 years ago

You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is

8_murik_8 [283]

Answer:

a = αR

Then We apply the Newton's second law of motion:

∑ F = ma

F - mg sinθ - f = ma.

F - mg sinθ - μmg cosθ = ma

Using given data we have:

F - 3217 = 470a ........................... (1)

Apply the Newton's law for rotation:

∑τ = I α

FR - (mg sinθ)R = $(\frac{mR^2}{2}+mR^2)\frac{a}{R}$

then: F - mg sinθ = 3ma / 2.

F - 2774 = 705a .......................... (2)

solving 1 and 2 we get:

F = 4103 N

a = 1.885 m/s²

b) In this case the time is given by:

t = $\sqrt{\frac{2d}{a}}$ it comes from: d = $v_ot+\frac{1}{2}at^2$ starting from rest vo = 0

t = $\sqrt{\frac{2*9}{1.885}$ = 3.09 s

8 0

3 years ago

What is force magnifier

fredd [130]

Answer:

It is an instrument which increases the amount of force available to a single hypothetical person.

Examples;

Levers, pulleys are force magnifiers.

6 0

3 years ago

A rock is dropped (from rest) off a bridge over the Merrimack River. The falling rock

rewona [7]

Answer:

31.25 meters or ~31 meters approximately

Explanation:

Let's see which of the 5 variables we are given since this is a constant acceleration problem.

$v_i \ \ \ \ \ \ t \\ v_f \ \ \ \ \ \triangle x \\ a$

We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.

We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².

We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.

Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, $v_i=0 \frac{m}{s}$ .

We want to find the vertical displacement, the height of the bridge, so we can say that $\triangle x= \ ?$

We have 4 out of 5 variables:

$v_i,\ a, \ t, \ \triangle x$

Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):

$x_f=x_i+v_it+\frac{1}{2}at^2$

Subtract $x_i$ from both sides of the equation to get:

$\triangle x=v_it+\frac{1}{2}at^2$

Substitute in our known variables and solve for delta x.

$\triangle x=(0\frac{m}{s})(2.5s) + \frac{1}{2} (-10\frac{m}{s^2})(2.5s)^2$

0 m/s multiplied by 2.5 s is 0, so we have:

$\triangle x =\frac{1}{2} (-10)(2.5)^2$

Evaluate the exponent first and multiply the terms together.

$\triangle x =(-5)(6.25)$
$\triangle x =-31.25$

The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.

7 0

3 years ago

Read 2 more answers

What is the mass of 5.4 moles of copper ?

Helen [10]

343.1484 grams. this is your answer.

3 0

4 years ago