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stiv31 [10]
3 years ago
13

Can someone help me with one question? “Do cells in the body help you grow?”

Physics
2 answers:
Gnoma [55]3 years ago
7 0

Answer: Body tissues grow by increasing the number of cells that make them up. Cells in many tissues in the body divide and grow very quickly between conception and adulthood.              yes

Explanation:

noname [10]3 years ago
5 0

Body tissues grow by increasing the number of cells that make them up. Cells in many tissues in the body divide and grow very quickly between conception and adulthood.
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45 j of 100 j of was used to turn on the television what is the percentage of energy used
Bingel [31]
45%, or 45/100 j. It's easy to find the percent if it's out of 100.
6 0
4 years ago
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Which term describes speed in a certain direction?
11Alexandr11 [23.1K]

Answer:

A. Velocity

Explanation:

Velocity is vector quantity thus has both magnitude and direction. It describes not only the speed but also the direction. Speed is scalar quantity so describes only speed but not direction. Energy has nothing to do with speed, acceleration describes change in velocity in a direction over time

8 0
2 years ago
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I need help with number 2 pls it’s a quiz grade
Eduardwww [97]
2 is correct hope you get it right
8 0
3 years ago
A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predete
Alex17521 [72]

Answer:

0.0366 m

Explanation:

We are given;

Current density; J = 540 A/cm² = 540 × 10⁴ m

Current; I = 0.57 A

Now, formula for current density is;

J = I/A

Where A is area = πr²

Thus;

J = I/(πr²)

r = √(I/(Jπ))

r = √(0.57/(540π))

r = 0.0183 m

Diameter = 2 × radius

Diameter = 2 × 0.0183

Diameter = 0.0366 m

7 0
3 years ago
A force in the x direction acts on a particle moving also in the x direction, producing a potential energy U(x)=Ax4 where A=0.63
poizon [28]

Answer:

The magnitude of the force is 1.29*10^-3N in the positive x direction

Explanation:

In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:

F(x)=-\frac{dU(x)}{dx}     (1)

where:

U(x)=Ax^4\\\\A=0.0630\frac{J}{m^4}

You replace the expression for U into the equation (1) and solve for F:

F(x)=-\frac{d}{dx}[Ax^4]=-4Ax^3     (2)

The force on the particle, for x = -0.080m is:

F=-4(0.630\frac{J}{m^4})(-0.0800m)^3=1.29*10^{-3}N

The magnitude of the force is 1.29*10^-3N in the positive x direction

7 0
3 years ago
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