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3 years ago

7

What if multiple forces act on an object in equilibrium??

1 answer:

tatuchka [14]3 years ago

4 0

Unless if all forces cancel each other out , the object will no longer be in equilibrium

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Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc

Paraphin [41]

Answer:

$V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}$

Explanation:

The 4 wires are connected in series: this means that the same current flow through them, and the voltage of the battery, V0, is equal to the sum of the voltages on each individual resistor:

$V_0=V_1+V_2+V_3+V_4$

Also, the equivalent resistance of the series circuit is

$R_{eq}=R_1+R_2+R_3+R_4$

The voltage V2 across wire 2 is given by Ohm's law:

$V_2 = R_2 I$ (1)

where I is the total current in the circuit, which is given by:

$I=\frac{V_0}{R_{eq}}=\frac{V_0}{R_1+R_2+R_3+R_4}$

Substituting this into eq. (1), we find an expression for V2:

$V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}$

8 0

3 years ago

A solid Cute hug sides 0.50m

irina [24]

Answer:

gg

Explanation:

hh

3 0

3 years ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg

lukranit [14]

Answer:

Mass of the steel cube = 7800 kg

Volume of the steel = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel = 7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

$Density = \frac{Mass}{Volume}$

We are given with density and sides of the cube

then volume of the cube

= $(side)^3$

= $10^3$

= 1000 cubic centimetre

Now

$7.8 = \frac{mass}{1000}$

$mass = 7.8 \times 1000$

mass = 7800 kg

<u>(2)volume of steel:</u>

Given the mass = 8 kg

$Density = \frac{Mass}{Volume}$

Substituting the values

$7.8 = \frac{8}{volume}$

$volume = \frac{8}{7.8}$

volume = 1.025 cubic centimetre

3 0

3 years ago