What are Earth's mineral resources and how long might<br> reserves last?

Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a

lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

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Explanation:

5 0

3 years ago

What is the mass of an object that requires a force of 182 N to accelerate at a rate of 13 m/s?

Inga [223]

Answer:

$m=14kg$

Explanation:

Hello.

In this case, since the force is defined in terms of the mass and acceleration by:

$F=m*a$

We can easily compute the mass by solving for it:

$m=\frac{F}{a}$

Whereas the force is 182 N (kg*m/s²) and the acceleration is 13 m/s², therefore, we obtain:

$m=\frac{182kg\frac{m}{s^2} }{13\frac{m}{s^2}}\\\\m=14kg$

Best regards.

6 0

3 years ago

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s

asambeis [7]

Answer:

$I=0.0987kg.m^2$

Explanation:

From the question we are told that:

Mass $M=1.80kg$

Deviation $d=0.250$

Time $t=0.940s$

Generally the equation for moment of inertia is mathematically given by

$I=\frac{T}{2\pi}^2(mgd)$

$I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)$

$I=0.0987kgm^2$

5 0

3 years ago

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

aksik [14]

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is $1 Mole=22.4L$

Given volume of rooms as $13.0ft\times12.0ft\times10ft=1560 ft^3$

Convert the volume into liters: $1560\times28.2l=43992l$

#From our conversion ratio above, we get the volume of air molecules in moles as:

$V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles$

Hence, the volume of air molecules is 1963.93 Moles

4 0

2 years ago

Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height

diamong [38]

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

Height of tree = 120,000 / 120

Height of tree = 1000

Hence, estimate height of tree = 1000 cm

5 0

2 years ago

What are Earth's mineral resources and how long might reserves last?