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Dmitry [639]
3 years ago
5

How long must a current of 100mA flow so as to transfer a charge of 80 C? ​

Physics
2 answers:
bulgar [2K]3 years ago
8 0

Answer:

800 secs

Explanation:

Q = it

80 = 100 x10^-3 t

t = 80/0.1

= 800secs

defon3 years ago
6 0

Answer:

800s

Explanation:

Q=It

where,

Q=80c

I =100MA=0.1A

t=?

Q=It

800=0.1×t

t=80×0.1

t=800s

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A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
During a supernova, the outer layers of a star are blown off and the star's core shrinks down by a factor of 10,000 or more to f
Lana71 [14]

Answer:

a. the core will spin faster.

Explanation:

By law of conservation of angular momentum

(mvR)i= (mvR)f

m= mass of star

v= speed of star

R= radius of star

i= initial

f= final

since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.

Hence, a. the core will spin faster.

8 0
3 years ago
What unit is used to measure the amount of energy used?
NNADVOKAT [17]

Answer: Joule

Explanation:

5 0
3 years ago
The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,
KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

3 0
3 years ago
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7
My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

4 0
2 years ago
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