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3 years ago

A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.

Physics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

$v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s$

Hence, the time interval of the marble is 6.09 seconds.

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How is reflection of light used in research

mel-nik [20]

In order to read the publications of his peers, or read his own notes of the work
that he did on the previous day, or find his coffee mug on his desk in the lab, the
research scientist must arrange to have each of them illuminated with visible
wavelengths of light, and then he must catch the light reflected from each of them
with his eyes.

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2 years ago

A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?

Artist 52 [7]

The formula for momentum is mass times velocity. Simply, we just multiply the given values:
p = mv
p = 40 kg x 4 m/s
p = 160 kg m/s

Other units for momentum is N s.
p = 160 N s

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2 years ago

two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist

Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

$R_{eq}=R_1+R_2$

For parallel combination,

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

When 6 ohm and 3 ohm are in series,

$R_s=6+3\\\\R_s=9\ \Omega$

When 6 ohm and 3 ohm are in paralle,

$\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega$

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0

2 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

1 year ago

When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.

Ivan

When an object in simple harmonic motion is at its maximum displacement, its acceleration is also at a maximum.

Reason: The speed is zero when the simple harmonic motion is at its maximum displacement, however, the acceleration is the rate of change of velocity. The velocity reverses the direction at that point therefore its rate of change is maximum at that moment. thus the acceleration is at its maximum at this point

Hope that helps!

7 0

2 years ago