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3 years ago

How much time is required for 48.96 C of charge to move through an electric juicer if the current through the juicer is 1.39 A?

1 answer:

8 0

The relationship between the charge flowing through a conductor, the current flowing through the conductor and the time is:
Q = It
Where Q is the charge, I is the current and t is the time of application of the current. Substituting the values:
48.96 = 1.39 x t
t = 35.2 seconds

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Which feature of the ocean floor lies just off the beach and is a part of continental crust

4vir4ik [10]

The coastline/shoreline
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2 years ago

Read 2 more answers

Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

Setler [38]

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

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5 0

2 years ago

You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.

ankoles [38]

Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

Mass = 60 Kg

Energy = 5000 J

We need to calculate the specific heat

Using formula of energy

$Q= mc\Delta T$

$c =\dfrac{Q}{m\Delta T}$

Put the value into the formula

$c=\dfrac{5000}{60\times(37-13)}$

$c=3.47222\ J/kg^{\circ}C$

Hence, The specific heat is 3.47222 J/kg°C.

5 0

2 years ago

3. Suppose that a block is pulled 16 meters across a floor. What amount of work is done if the force used to drag the block is 2

Nadusha1986 [10]

Explanation:

Work is the product of force and distance.

W = F×d

W = (22 N) × (16 m)

W = 352 J

7 0

3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

ValentinkaMS [17]

Answer:

$2.02672\times 10^{-11}\ V$

Explanation:

f = Frequency = 75 Hz

B = Magnetic field = $1\times 10^{-3}\ T$

d = Diameter of cell = $7.4\ \mu m$

t = Time taken

$\omega$ = Angular frequency

The expression of emf is

$E=NAB\omega sin\omega t$

Emf is maximum when $sin\omega t=1$ and N = number of turns = 1

$E_m=AB\omega\\\Rightarrow E_m=\pi \left(\frac{d}{2}\right)^2 \times B \times 2\pi\times f\\\Rightarrow E_m=\pi \left(\frac{7.4\times 10^{-6}}{2}\right)^2\times 1\times 10^{-3}\times 2\pi\times 75\\\Rightarrow E_m=2.02672\times 10^{-11}\ V$

The maximum emf that can be generated around the perimeter of a cell in this field is $2.02672\times 10^{-11}\ V$

4 0

2 years ago