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julsineya [31]
3 years ago
7

Picturing the way a volcano erupts to remember how a volcano functions is called a. Visualizing c. Keywording b. Categorizing d.

All of these
Physics
2 answers:
xenn [34]3 years ago
8 0
Definitely a visualizing
Yuliya22 [10]3 years ago
7 0
The answer is A visual 
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7. A motorcycle accelerates from rest at a rate of 4 m/s2 while traveling 60m. What is the motorcycle's velocity at
Daniel [21]

Answer: C

Explanation: 60 divided by 4 =15

6 0
3 years ago
Suppose, we have a parallel plate capacitor and we know the following about it: Area of each plate = $0.0012m^2$ Distance betwee
FromTheMoon [43]

Answer:

udegsgsfsjtajtsiaitskhsigsfjsjgsyoeyodogdoydyosiskgoydoydiydiyddohdohdohyodhodohdoydoydoydosoydogdkgdohdhdohdgodyodoydoydyodoydyodoydohxohdohxohxohxohxogxkx skzjgz a few days

7 0
4 years ago
Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex
Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, \rho=1.68\times 10^{-8}\ \Omega-m

We need to find the resistance of the copper wire. It is given by :

R=\rho\dfrac{l}{A}

R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0
3 years ago
I would like to know why this is the correct answer
Helen [10]

The acceleration of the object if the net force is decreased = 0.13 m/s²

<h3>Further explanation</h3>

Given

A net force of 0.8 N acting on a 1.5-kg mass.

The net force is decreased to 0.2 N

Required

The acceleration of the object if the net force is decreased

Solution

Newton's 2nd law :

\tt \sum F=m.a

The mass used in state 1 and 2 remains the same, at 1.5 kg

  • state 1

ΣF=0.8 N

m=1.5 kg

The acceleration, a:

\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.8}{1.5}\\\\a=0.53`m/s^2

  • state 2

ΣF=0.2 N

m=1.5 kg

The acceleration, a:

\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{0.2}{1.5}\\\\a=0.13~m/s^2

8 0
3 years ago
When a force of 450N pushes on a 20kg box as
ehidna [41]

Answer:

ma+mgsinh0​+f=F∴(25)(0.75)+(25)(10)sinh0​+μk​N=F∴18.75+(250)(0.6h)+μk​(mgcosh0​=F⟹18.75+150+μk​((25)(10)(0.76))=500∴168.75+μk​(190)=500⟹μk​(190)=331.25⟹μk​=1.74

Explanation:

7 0
3 years ago
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