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sladkih [1.3K]
2 years ago
14

Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

ar and apply forces to it. Use the format of "if . . . then . . . because . . .” and be sure to answer the lesson question "How can Newton's laws be experimentally verified?” specific to Newton’s second law.
Physics
2 answers:
elena55 [62]2 years ago
8 0

For e2020 this is the simple answer "If force is applied to a car, then its acceleration will change proportionally, as predicted by Newton’s second law, F = ma."

{Just did this} But I used "If the toy car has force applied to it then the force would leave no choice but to move the toy car in whichever direction it is applied because the force would move the toy car according to Newton's laws." and got it right.

Tanzania [10]2 years ago
5 0
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.

How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.

Hope this helps.
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Answer:

Explanation:

delta V = v * alpha * delta T

= V * 0.00053 * (92.2 - 55.0)

= 0.019716 V

percentage that the owner

= [delta V / V] * 100

= [0.019716 V / V] * 100

= 1.9716 %

4 0
2 years ago
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the frequency of the waves is doubled
m_a_m_a [10]

Answer:

halved

Explanation:

The velocity of the a wave is obtained by multiplying the frequency and wavelength.

v=f\lambda\\\Rightarrow f=\frac{v}{\lambda}\\\Rightarrow \lambda=\frac{v}{f}

Where

v = Velocity

f = Frequency

\lambda = Wavelength

The velocity here is constant. So, if the frequency is doubled the wavelength is halved.

6 0
2 years ago
A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the ro
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To solve this problem it is necessary to take into account the concepts related to Centripetal Force and Friction Force.

In the case of the centripetal force, we know that it is defined as

F_c = \frac{mv^2}{R}

Where,

m=mass

v= velocity

r= Radius

In the case of the Force of Friction we have to,

F_f = \mu m*g

Where,

\mu =Friction Constant

m= mass

g= gravity

According to the information given, the centripetal force must be less than or equal to the friction force to stay on the road, in this way

\frac{mv^2}{R} \leq \mu m*g

Re-arrange to find the velocity,

v \leq \sqrt{\mu gR}

v \leq \sqrt{(0.6)(9.8)(300)}

v \leq 42m/s

Therefore la velocidad del carro debe ser igual o menor a 42m/s para mantenerse en el camino

6 0
2 years ago
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Flura [38]

Answer:

the correct answer is A, the object goes 4 times as far

Explanation:

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        R1 = v₀² sin 2θ / g

Now let's double vo, the new speed is

         v = 2 v₀

We calculate the scope

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         R2 = 4 v₀² sin 2θ / g

         R2 = 4 R1

Therefore the correct answer is A, the object goes 4 times further

5 0
2 years ago
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