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Jet001 [13]
3 years ago
8

What weighs exactly 500 grams in the house?

Physics
1 answer:
Serga [27]3 years ago
3 0
Another common household items is AA Alkaline batteries , I found these weigh approx 23 gram of them would have a mass of around 500 +/-5%. Never or unused , same mass
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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
2 years ago
A magnetic field would be produced by a beam of
kumpel [21]
3 protons should be your answer
4 0
2 years ago
4. How does nearness to water affect the climate zone of an area?
Afina-wow [57]

Answer:

Choice A.

Nearness to a body of water causes an increase in humidity, due to the higher rate of evaporation.

6 0
2 years ago
Read 2 more answers
an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object
emmainna [20.7K]

Answer:

The acceleration of the object is -30\ m/s^2

Explanation:

Given:

Initial velocity of object v_i = 200 feet/second

Final velocity of object v_f = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration a is given by:

a=\frac{v_f-v_i}{t}

where v_f represents final velocity, v_i represents initial velocity and t is time of travel.

Plugging in values to evaluate acceleration.

a=\frac{50-200}{5}

a=\frac{-150}{5}

a=-30\ m/s^2

The acceleration of the object is -30\ m/s^2 (Answer). The negative sign shows the object is slowing down.

4 0
3 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
2 years ago
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