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vampirchik [111]

3 years ago

14

A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of f

riction between the box and the floor is Î¼. At what rate does the child do work on the box?

1 answer:

dexar [7]3 years ago

4 0

Answer:

Explanation:

Given

mass of box is m

speed of box is v

coefficient of friction $\mu =\frac{1}{4}=0.25$

Box is moving at a constant speed i.e. Net Force on box is zero

i.e. Frictional Force and Force Applied are Equal

$F=f_r=\mu mg$

The Rate at which power is generated is

$P=F\cdot v$

$P=\mu mg\cdot v$

$P=0.25mgv$

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A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th

zheka24 [161]

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

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2 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

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4 years ago

A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?

WITCHER [35]

Answer:

b

Explanation:

3 0

3 years ago

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