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3 years ago

7

Two objects are attracted to each other by electromagnetic forces. Suppose I double the charge of one object. How can I return t

he force to its original value?

1 answer:

nata0808 [166]3 years ago

4 0

Answer:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

Explanation:

Lets assume,

Charge on first object = Q

Charge on second object = q

Distance between them = r

Force between the two charges = F

According to Coulomb's law,

$F = k \frac{Qq}{r^{2}}$

where, k = Coulomb constant

New value of charge on first object = 2Q. Thus the new force(F') will be

$F' = k \frac{2Qq}{r^{2}}$

$F' = 2F$

So, to bring the value of force(F') to original value, there are two options:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

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How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$

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4 years ago

Suppose you ride your bike to the library traveling at 0.5 km/min. It takes you 25 minutes to get to the library. How far did yo

madam [21]

Answer= 12.5 km

0.5 km x km
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4 0

3 years ago

Read 2 more answers

Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Molodets [167]

Answer:

r = 5.335 meters

Explanation:

Given that,

Charge 1, $q_1=-165\ \mu C$

Charge 2, $q_2=115\ \mu C$

Force of attraction between two charges, F = 6 N

The force of attraction between two charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$ , r is the separation between two charges

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 165\times 10^{-6}\times 115\times 10^{-6}}{6}}$

r = 5.335 m

So, the separation between two charges is 5.335 meters. Hence, this is the required solution.

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3 years ago