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OverLord2011 [107]
2 years ago
12

A particle with a charge of 4.4 * 10^-5 C is released in an electric field whose magnitude is 750 N/C. What is the force that th

e field exerts on the particle?
Physics
1 answer:
padilas [110]2 years ago
4 0

Answer:

F = 3.3×10^ -2 N

.................

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Describe in your own words what is meant by the statement that the Sun is in hydrostatic equilibrium.
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Answer and Explanation:

Hydrostatic equilibrium is the condition in which force is balance that is upward force and downward force the downward force is due to gravitational force and the upward force is due to the pressure. The Sun is said to be in hydrostatic equilibrium means the force acting on it is balance means upward force which is due to pressure is same as the force exerted by gravitation.

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What are the three main branches of science
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What os environment?​
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is the environment in which users run application software. The environment consists of a user interface provided by an applications manager and usually an application programming interface (API) to the applications manager.

7 0
3 years ago
A simple cyanine dye molecule has a long central carbon chain, with a total of 8 bonds, between the two nitrogen atoms. One elec
wariber [46]

Answer:

For one dimensional box

E= n^{2}h^{2}/ 8mL^{2}

For lowest energy transition will be from n =1 to n=2

Hence,

hc/\lambda = 2^{2}h^{2}/ 8mL^{2} - h^{2}/8mL^{2}

h= 6.626 *10-34 J.s

m = 9.31 *10-31 Kg

L= 0.14 * 10-9 m

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\lambda = 8*c*m*L^{2} / 3*h

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4 0
3 years ago
College Physics Homework Please help MathPhys man i need ur hwlp
pogonyaev

Answer:

5.9 m/s²

78 N

Explanation:

Draw a free body diagram for each mass.

The mass on the left (M) has a tension force pulling up (T) and a weight force pulling down (Mg).

The mass on the right (m) has a tension force pulling up (T) and a weight force pulling down (mg).

Apply Newton's second law to the mass on the left (remember it accelerates down):

∑F = ma

T − Mg = M (-a)

T − Mg = -Ma

Apply Newton's second law to the mass on the right:

∑F = ma

T − mg = ma

Two equations, two unknowns (T and a).  First, we want to find a.  So start by subtracting the first equation from the second equation.

(T − mg) − (T − Mg) = ma − (-Ma)

T − mg − T + Mg = ma + Ma

Mg − mg = (m + M) a

a = g (M − m) / (m + M)

Given M = 20 kg and m = 5 kg:

a = 9.8 m/s² (20 kg − 5 kg) / (5 kg + 20 kg)

a = 5.88 m/s²

Now plug into either equation to find the tension.

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5 kg) (9.8 m/s² + 5.88 m/s²)

T = 78.4 N

Rounding to two significant figures, the acceleration and tension are 5.9 m/s² and 78 N.

Notice that the radius of the pulley and the distance between the masses were extra information.

4 0
3 years ago
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