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Sholpan [36]
2 years ago
8

A 150 g egg is dropped from 3.0 meters. The egg is

Physics
1 answer:
Lynna [10]2 years ago
5 0

<u><em>Answer:</em></u>

<u><em> </em></u>

<u><em>9.2 N, with significant figure rounding (2 s.f.) </em></u>

<u><em></em></u>

<u><em>Explanation:</em></u>

<u><em>This problem can be solved using momentum. The following equation relates momentum (mass & velocity) with force and time:</em></u>

<u><em></em></u>

<u><em>Note that  where v is the final velocity and v₀ is the initial velocity. Δv just means change in velocity.</em></u>

<u><em></em></u>

<u><em>Mass of the egg is 150 g, but we need to convert to kilograms if we want to use Newtons as a unit. 150 g is equal to 0.15 kg. since 1000 g = 1kg. </em></u>

<u><em>m = 0.15 kg</em></u>

<u><em></em></u>

<u><em>The dropped from 3.0 meters is irrelevant as the question tells us the initial velocity of the egg: 4.4 m/s before it hits the ground.</em></u>

<u><em>v₀ = 4.4 m/s [down]</em></u>

<u><em></em></u>

<u><em>When it comes to a stop, the egg will have a velocity of 0.</em></u>

<u><em>v = 0 m/s</em></u>

<u><em></em></u>

<u><em>The time it takes for the egg to stop is 0.072 seconds.</em></u>

<u><em>Δt = 0.072 s</em></u>

<u><em></em></u>

<u><em>Therefore, if down is positive, then</em></u>

<u><em></em></u>

<u><em>   </em></u>

<u><em></em></u>

<u><em>We round to two significant figures since every quantity has two sig. figs.</em></u>

<u><em>We only care about the magnitude, not direction. The answer is 9.2 N.</em></u>

<u><em>Unlimited, ad-free access to all of the questions with Brainly Plus</em></u>

<u><em>START 7 DAY FREE TRIAL</em></u>

<u><em>Click to let others know, how helpful is it</em></u>

<u><em>5.0</em></u>

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If two normal distributions each have the same mean, but one has a standard deviation of 1 and the other has a standard deviatio
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A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ
lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC  at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

U=\frac{Kq_{1}q_{2}}{r}

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

U = U_{1,2}+U_{2,3}+U_{3,1}

U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}

U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}

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2 years ago
The terminals of a battery are connected across two resistors in series. The resistances of the resistors are not the same. Whic
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Answer:

(c) The current in each resistor is the same.

Explanation:

When two resistors are connected in series, we have the following:

- The resistors are connected such that the current passing through the two resistors is the same

- The voltage of the battery is equal to the sum of the voltage drops across each resistor

- the equivalent resistance of the circuit is equal to the sum of the individual resistances:

R = R1 + R2

So, let's analyze each statement:

(a) The resistor with the smaller resistance carries more current than the other resistor. --> FALSE. The current through the two resistors is the same.

(b) The resistor with the larger resistance carries less current than the other resistor. --> FALSE. The current through the two resistors is the same.

(c) The current in each resistor is the same. --> TRUE.

(d) The potential difference across each resistor is the same. --> FALSE: the potential difference across each resistor is given by

V=RI

where I (the current) is the same for both resistors, while R (the resistance) is not, so V is also different for the two resistors.

(e) The potential difference is greatest across the resistor closest to the positive terminal --> FALSE. According to

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4 0
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A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle i
olga55 [171]

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

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1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

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1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

6 0
3 years ago
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