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Lunna [17]
3 years ago
6

How much charge passes through a wire in 4.0 s if the current is 3.0 A?

Physics
1 answer:
NemiM [27]3 years ago
6 0

Answer:

12 coloumbs

Explanation:

Q = IT

I= 3A, T=4s

Q = 3×4 =12coloumbs

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The half-life of iodine-131 is 8.04 days. It’s decay constant equals
Svetllana [295]

Answer:

Explanation:

we know that half life of an element is

T=0.693/λ

where λ is decay constant in order to find decay constant

λ=0.693/T

λ=0.693/8.04

λ=0.086

7 0
3 years ago
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
3 years ago
A swimmer swims at 5 m/s. How long would it take to swim 5 laps of a 50m pool?​
Gre4nikov [31]
About 5 hours gooood luck
3 0
1 year ago
Pls Help Me with this problem
Naddika [18.5K]

Answer:

explanation

Explanation:

1 = C

2 = A

3 = D

4 = E

5 = B

8 0
3 years ago
Read 2 more answers
You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
3 years ago
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