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2 years ago

We eat pizza from the inside out???

Physics

2 answers:

lara [203]2 years ago

6 0

Answer:

what

Explanation:

what I dont get it??

solong [7]2 years ago

5 0

Answer:

that makes me confused I don't think of pizza the same again LOL

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What is 34 + (5) × (1.2465) written with the correct number of significant figures?

bonufazy [111]

Answer: 40

Explanation:

= 34 + 5 * 1.2465

= ‭40.2325‬

= 40

The number of significant figures in the answer should be the same as the number with the least number of significant figures that any of the digits in the equation have.

32 has 2 significant figures so the answer has to be 2 significant figures which is 40.

7 0

3 years ago

a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr

Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

$f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}}$ where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

$f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz$

5 0

2 years ago

What’s the answer for this problem?

pickupchik [31]

The answer is always true a

4 0

2 years ago

_____ help by supporting the plant and holding the leaves up to the light.

Semenov [28]

The answer is the stem.

7 0

3 years ago

Read 2 more answers

The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?

elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, $B=2.7\ mT=2.7\times 10^{-3}\ T$

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

$B=\dfrac{\mu_o I}{2r}$

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}$

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}$

r = 0.00238 m

$r=2.38\times 10^{-3}\ m$

Hence, this is the required solution.

8 0

2 years ago