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3 years ago

A heavy truck and a small truck roll down a hill. Neglecting friction, at the bottom of the hill the heavy truck has greater

Physics

1 answer:

postnew [5]3 years ago

3 0

Answer:

kenetic energy

Explanation:

or potential energy

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PART ONE

Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass. Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

$\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right$

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0

3 years ago

A transformer consists of 290 primary windings and 824 secondary windings. Part A: If the potential difference across the primar

jasenka [17]

Answer:

Part 1) Voltage in secondary windings is 61.08 Volts

Part 2) Current in secondary windings is 0.53 Amperes

Explanation:

The potential developed in the primary and secondary winding of a transformer are related as

$\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}$

where

Np no of turns in primary coil

Ns no of turns in secondary coil

Vp Voltage of turns in primary coil

Vs Voltage of turns in secondary coil

Applying values in the formula we get

$\frac{290}{824}=\frac{21.5}{V_{s}}\\\\\therefore V_{s}=21.5\times \frac{824}{290}=61.08V$

Part 2)

Using Ohm's law the current is given by

$I=\frac{V_{s}}{R}\\\\I=\frac{61.089}{115}=0.53A$

5 0

2 years ago

PLEASE HELP ME ANSWER DUE SOON BRAINLIEST FOR WHOEVER

DerKrebs [107]

Answer:

1) mechanical

2) transverse

3) medium

4) longitudinal

Explanation:

7 0

2 years ago

The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these tw

timurjin [86]

Answer:

$1.09527\times 10^{-8}\ N$

Explanation:

$q_1$ = Mg ion = $+2q$

$q_2$ = F ion = $-q$

q = Charge of electron = $1.6\times 10^{-19}\ C$

r = Distance between ions = $0.072+0.133\ nm$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electrical force is given by

$F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N$

The attractive force is $1.09527\times 10^{-8}\ N$

8 0

3 years ago

You are driving at 75 km/h. Your sister follows in the car behind at 75 km/h. When you honk your horn, your sister hears a frequ

OLga [1]

The ideal concept for solving this question is based on the Doppler effect, for which it is indicated that the source's listening frequency changes as the distance and the relative speed between the receiver and the transmitter are also changed. However, if the relative velocity between the two objects is zero as in the particular case presented (since both travel at 75km / h) we have that there will be no change in frequency.

Therefore the frequency that I hear and that my sister would listen would be the same.

5 0

3 years ago