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3 years ago

Which of the following could be considered a scientific statement?

Physics

2 answers:

tresset_1 [31]3 years ago

5 0

B. Nancy is the tallest person in the class.

aivan3 [116]3 years ago

3 0

B as it isn’t a opinion instead it’s a fact

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What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a

Alex777 [14]

Answer: d. 5 m/s^2

Explanation:

Acceleration is the change in velocity in a given time.

a = (30-20)/2 = 5

7 0

3 years ago

Read 2 more answers

A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.

alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Here,

k = Spring constant

m = Mass

Our values are given as,

$m = 35g = 35*10^{-3}kg$

$f = 1 Hz$

Rearranging to find the spring constant we have that,

$k = (2\pi f \sqrt{m})^2$

$k = 4\pi^2 f^2 m$

$k = (4) (\pi)^2 (1) (35*10^{-3})$

$k = 1.38N/m$

Therefore the spring constant is 1.38N/m

7 0

4 years ago

A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti

Alenkinab [10]

Answer:

$v_f = 3.45 m/s$

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

$mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 10 kg

$\theta = 10^0$

L = 3.5 m

so we will have

$10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)$

$2(9.81) sin10 \times 3.5 = v_f^2$

so final speed is given as

$v_f = 3.45 m/s$

6 0

3 years ago

A capacitance C and an inductance L are operated at the same angular frequency.

Levart [38]

A) $\omega = \frac{1}{\sqrt{LC}}$

The magnitude of the capacitive reactance is given by

$X_C = \frac{1}{\omega C}$

where

$\omega$ is the angular frequency

C is the capacitance

While the magnitude of the inductive capacitance is given by

$X_L = \omega L$

where L is the inductance.

Since we want the two reactances to be equal, we have

$X_C = X_L$

So we find

$\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}$

B) 7449 rad/s

In this case, we have

$L=5.30 mH = 5.3\cdot 10^{-3}H$ is the inductance

$C= 3.40 \mu F= 3.40 \cdot 10^{-6}F$ is the capacitance

Therefore, substituting in the formula for the angular frequency, we find

$\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s$

C) $39.5 \Omega$

Now we can us the formulas of the reactances written in part A). We have:

- Capacitive reactance:

$X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega$

- Inductive reactance:

$X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega$

7 0

3 years ago

Please explain your answer when you found the answer (Only question number 9)

Zinaida [17]

<h3>With a uniform acceleration of 2 m/s ²</h3>

5 0

3 years ago