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Nady [450]
3 years ago
9

For the long life cells we have to connect them in ____ combination​

Physics
2 answers:
NISA [10]3 years ago
5 0

Answer:

Parallel combination.

Grace [21]3 years ago
3 0

for the long life cells we have to connect them in <em><u>parallel</u></em> combination

hope it is helpful to you

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I need help in physics please help me
artcher [175]

The first step is to represent the vectors shown in the image in Cartesian coordinates.

For the vector C we have a magnitude of 4.8 and an angle 22 ° with the axis -y (direction j)

To write this vector in Cartesian coordinates we must find its component in x (address i) and in the y axis.

x = 4.8sin(22)i\\\\y = 4.8cos(22)(-j)

So:

C = 4.8sin(22) i + 4.8cos(22)(-j)\\\\C = 1.798\ i - 4.450\ j

For Vector B we have a magnitude of 5.6 and an angle of 33 with the -x axis (-i direction)

So:

x = 5.6cos(33)(-i)\\\\y = 5.6 sin(33)(-j)

So:

B = 5.6cos(33)(-i) + 5.6sin(33)(-j)\\\\B = -4.696\ i - 3.05\ j

Finally the sum of B + C is made component by component in the following way:

F = (-4.696 +1.798)i + (-4.450 - 3.05)j\\\\F = -2.898\ i - 7.5\ j

Finally the magnitude of f is:

|F| = \sqrt{(-2,898)^2 + (-7.5)^2}

| F | = 8.04

3 0
4 years ago
Use the formula from Newtons second law, namely Fnet=∆t=∆P that the Net force exerted on the object is equal to the product of t
Aleksandr-060686 [28]

Answer:

Explanation:

Fnet = Δt = ΔP

Fnet = ΔP/Δt

= mv - mu / t

= m(v - u) / t

But we know v = u + at

=> a = v - u / t

Therefore,

Fnet = ma

6 0
2 years ago
In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi
Ronch [10]

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

6 0
3 years ago
A jet airplane has a velocity of 1145 knots. A knot is 1 nautical mile (nm)/hr. A nautical
Tanya [424]

Answer:

1 m = 39.37 in = 39.37/12 ft = 3.28 ft

V = 1145 k/hr  = 1145k/hr * 6076 ft/k = 6957020 ft / hr

V = 6957020 ft/hr / 3600 s/hr = 1933 ft/sec

V = 1933 ft/sec / (3.28 ft / m) = 589 m/s

Check:

88 ft/sec = 60 mph

(1145 k/hr * 6076 ft / k) 3600 sec/hr = 1933 ft/sec = 589 m/s

1933 ft/sec / (88 ft/sec) * 60 mph = 1318 mph

Also,  1318 / 1145 = 6076 / 5280       as it should

4 0
3 years ago
A branch falls from a tree How fast is the branch moving after 0 28 seconds
egoroff_w [7]

Answer:

c. 2.7 m/s

Explanation:

v = a.t + v_{0}\\a = g = 9.81 m/s^{2}} \\t=0.28 s \\v_{0} = 0\\=> v = 9.81 * 0.28 = 2.74 m/s

3 0
3 years ago
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