Please help me i have this due tommorrow!!!

A student observes that for the same net force heavier objects accelerate less which statement describes the correct conclusion?

rodikova [14]

Answer: A decrease in acceleration causes the mass to increase.

Explanation:

Let's rewrite this as:

"For a constant force F, we can see that heavier objects accelerate less"

Then, as the mass increases, the acceleration of the object decreases.

This is known as an inverse relationship between acceleration and mass.

By second Newton's law we have that:

F = m*a

Force equals mass times acceleration.

So if the force is constant, we can write it as:

a = F/m.

In this equation, we can see that, if the acceleration decreases and F remains constant, then the mass m must increase (because it is in the denominator)

Then the correct option is:

A decrease in acceleration causes the mass to increase.

4 0

3 years ago

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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0

4 years ago

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The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column

JulsSmile [24]

Answer:

1) f = 214 Hz , 2) answer is c , 3) f = 428 Hz , 4) f₂ = 428 Hz , f₃ = 643Hz

Explanation:

1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore

L = λ / 2

λ = 2L

λ = 2 0.8

λ = 1.6 m

wavelength and frequency are related to the speed of sound (v = 343 m / s)

v =λ f

f = v / λ

f = 343 / 1.6

f = 214 Hz

2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced

λ' = 2 L ’

as L ’<L₀

λ' <λ₀

f = v / λ'

f' > fo

the correct answer is c

3) in this case the length is L = 0.40 m

λ = 2 0.4 = 0.8 m

f = 343 / 0.8

f = 428 Hz

4) the different harmonics are described by the expression

λ = 2L / n n = 1, 2, 3

λ₂ = L

f₂ = 343 / 0.8

f₂ = 428 Hz

λ₃ = 2 0.8 / 3

λ₃ = 0.533 m

f₃ = 343 / 0.533

f₃ = 643 Hz

4,1) as we have two maximums at the ends, all integer multiples are present

the answer is C

E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested

in this pipe there is a maximum in the open part and a node in the closed part, so the expression

L = λ / 4

L = 1.6 / 4

L = 0.4 m

the answer is C

F) in this type of pipe the general expression is

λ = 4L / n n = 1, 3, 5 (2n + 1)

therefore only odd values can produce standing waves

λ₃ = 4L / 3

λ₃ = 4 0.4 / 3

λ₃ = 0.533

f₃ = 343 / 0.533

f₃ = 643 Hz

5 0

4 years ago

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

"The toy car is about 3 inches long" is an example of a what observation?

Crazy boy [7]

Answer:

It's a quantitative observation because it includes numerical data.

Explanation:

please mark brainliest if im correct

5 0

4 years ago