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klasskru [66]
3 years ago
10

A -0.06 charge that moves downward is in a uniform electric field with a strengthened of 200 N/C. What is the magnitude and dire

ction of the force on the charge
Physics
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

The magnitude of the force is 12 N Upwards

Explanation:

The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.

Given;

magnitude of charge, q = 0.06 C

magnitude of electric field, E = 200 N/C

The magnitude of the force is given by;

F = qE

F = 0.06 x 200 N/C

F = 12 N Upwards

Therefore, the magnitude of the force is 12 N Upwards

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Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
Mariulka [41]

Answer:

a) 0.0288 grams

b) 2.6*10^{-10} J/kg

Explanation:

Given that:

A typical human  body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance  for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = \frac{energy \ absorbed }{mass \ of \ the \ body}

= \frac{1.10*10^6*1.6*10^{-14}}{80}

= 2.2*10^{-10} \ J/kg

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) =  1.2  *2.2*10^{-10} \ J/kg

Effective dose (Sv) = 2.6*10^{-10} J/kg

 

5 0
3 years ago
Two children ride side-by-side on a carousel. Their paths are shown in the image below.
Sonbull [250]

Answer:

The child represented by a star on the outside path.

Explanation:

5 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

So final speed \frac{3.49\times 75}{100}=2.6175rad/sec

Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

(a) Now final velocity \omega =0rad/sec

So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

(b) The work done in coming rest is given by

\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J

4 0
2 years ago
In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on
vekshin1

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

3 0
2 years ago
The 10-kg uniform rod is pinned at end
Anton [14]
Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left 
Sum moments about the pivot to zero. 
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0 
by iterative answer we discover that 
θ ≈ 0.465 radians 
θ ≈ 26.6º 
3 0
3 years ago
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