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nikdorinn [45]
3 years ago
15

What is the strength of the electric field of a point charge of magnitude +4.8 × 10^-19 C at a distance of 4.0 × 10^-3 m? A. 3.6

× 10^-4 N/C B. -2.7 × 10^-4 N/C C. -3.6 × 10^-4 N/C D. 2.7 × 10^-4 N/C
Physics
1 answer:
soldi70 [24.7K]3 years ago
8 0
Answer is D -  from the equation of electric field for point charge
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The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration​
Veseljchak [2.6K]

Answer:

acceleration = v-u/ t

= 15-10/5

= 5/5

= 1 m/s2

Explanation:

hope this helped you.

7 0
3 years ago
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Where does the cell store DNA and RNA
Anarel [89]
DNA is found mainly in the nucleus of the cell, while RNA is found mainly in the cytoplasm of the cell although it is usually synthesized in the nucleus.
7 0
3 years ago
A circular loop of wire with radius 2.00 cm and resistance 0.600 Ω is in a region of a spatially uniform magnetic field B⃗ that
Anika [276]

Answer:

Incomplete questions

Let assume we are asked to find

Calculate the induced emf in the coil at any time, let say t=2

And induced current

Explanation:

Flux is given as

Φ=NAB

Where

N is number of turn, N=1

A=area=πr²

Since r=2cm=0.02

A=π(0.02)²=0.001257m²

B=magnetic field

B(t)=Bo•e−t/τ,

Where Bo=3T

τ=0.5s

B(t)=3e(−t/0.5)

B(t)=3exp(-2t)

Therefore

Φ=NAB

Φ=0.001257×3•exp(-2t)

Φ=0.00377exp(-2t)

Now,

Induce emf is given as

E= - dΦ/dt

E= - 0.00377×-2 exp(-2t)

E=0.00754exp(-2t)

At t=2

E=0.00754exp(-4)

E=0.000138V

E=0.138mV

b. Induce current

From ohms laws

V=iR

Given that R=0.6Ω

i=V/R

i=0.000138/0.6

i=0.00023A

i=0.23mA

5 0
3 years ago
Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of
Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

A=2\times10^{-6}m^2

* The young modulus of the steel is,

Y=2.1\times10^{11}\text{ Pa}

Solution:

The young modulus of the steel in terms of the force and extension is,

Y=\frac{F\times l}{A\times dl}

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0
1 year ago
A racing car has a mass of 1720 kg. the acceleration of gravity is 9.8 m/s 2 . what is its kinetic energy if it has a speed of 1
lesantik [10]
Is this to late to answer?

3 0
3 years ago
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