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2 years ago

Two negative charges that are both -0.3C push each other apart with a force of 19.2 N. How far apart are the two charges?

Physics

1 answer:

Paha777 [63]2 years ago

8 0

Using Coulomb's law: k*(-0.3)*(-0.3)/(d^2)=19.2 D is the distance between the two negative charges

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For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

d = 8 furlongs

It traveled 8 furlongs for the first 4.0 seconds.

8 0

2 years ago

A microwave oven operates at 2.60 ghz . what is the wavelength of the radiation produced by this appliance?

cestrela7 [59]

10.3 cm The wavelength will be the distance that light travels in 1 second divided by the frequency of the radiation. Since the over operates at 2.60 ghz, the frequence is 2.6 billion times per second, or 2.60 x 10^9. The speed of light is defined as 299792458 m/s exactly. So 299792458 m/s / 2.60 x 10^9 1/s = 0.10337671 m = 10.337671 cm Since we only have 3 significant digits, the answer rounds to 10.3 cm

8 0

2 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across

Rina8888 [55]

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

8 0

2 years ago