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2 years ago

Find the mass of a child who runs at a speed of 4 m/s to get a pizza with extra cheese. His momentum is 120 kg•m/s.

1 answer:

7 0

Wow! Very simple. Use the equation for momentum which is p=mv. You know your momentum and velocity, so do 120=m(4), now solve for m, which is 30. The mass of the child is 30 kg

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What describes the particles in a liquid

Gnoma [55]

Answer:

liquid a particles slides past pother

Explanation:

mark brainliest :))

8 0

3 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

5 0

3 years ago

A car is traveling at a speed of 10m/s. A 0.5kg clump of mudis

CaHeK987 [17]

Answer:B

Explanation:

Given

speed of car $v=10 m/s$

mass of clump $m=0.5 kg$

Radius of car tire $r=0.2 m$

Since the tire is rotating about axle so a centripetal force is acting constantly on each particle towards the center of tire.

Centripetal force is given by

$F_c=\frac{mv^2}{r}$

where $m=mass\ of\ element$

$v=speed$

$r=distance\ from\ center$

$F_c=\frac{0.5\times 10^2}{0.2}$

$F_c=250\ N$ (inward)

3 0

3 years ago

A bird lands on a bare copper wire carrying a current of 51

nirvana33 [79]

The resistance of the piece of wire is
$R= \frac{\rho L}{A}$
where
$\rho = 1.68 \cdot 10^{-8}\Omega m$ is the resistivity of the copper
$L=5.1 cm=0.051 m$ is the length of the piece of wire
$A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2$ is the cross sectional area of the wire
By substituting these values, we find the value of R:
$R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega$

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
$V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V$

7 0

3 years ago

ILL MARK BRAINLIEST <br>PLEASE HELP <br>2.Write the formula associating energy with power

bekas [8.4K]

Answer:

The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.

Explanation:

Sub to Beast_Building on yt

7 0

2 years ago