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aev [14]
3 years ago
9

8TH GRADE SCIENCE I NEED NOW DO NOT SKIP

Physics
1 answer:
yan [13]3 years ago
7 0

Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

=50(7.8)= 390N

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Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST
Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

a^2+b^2=c^2

Plug values in

88.4^2+174^2=c^2

Simplify

7814.56+30276=c^2

Add the two values

38090.56=c^2

Take the square root of both sides

195.168\approx195.168

8 0
2 years ago
Is it possible to calculate the displacement based on an elapsed time from a position time graph?
Nataly [62]
No I don’t think so. But it worth a try tho. Try it out.
5 0
3 years ago
Which of the following is true about conduction? Question 2 options: It is heat transfer through direct contact It can only occu
kramer

Answer:

through direct contact

Explanation:

took the test!

3 0
3 years ago
Hey everyone<br><br><br>what is optics??​
lana66690 [7]

Explanation:

Optics is a branch of physics that is the study of light and vision. ... The branch of physics dealing with the nature and properties of electromagnetic energy in the light spectrum and the phenomena of vision. In the broadest sense, optics deals with infrared light, visible light, and ultraviolet light.

7 0
3 years ago
For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103
Alona [7]

Answer:

a) P = 44850 N

b) \delta l =0.254\ mm

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

\sigma=\frac{\textup{Load}}{\textup{Area}}

on substituting the values, we get

345\times10^6=\frac{\textup{Load}}{0.00013}

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation (\delta l) due to an axial load is given as:

\delta l =\frac{PL}{AE}

on substituting the values, we get

\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}

or

\delta l =0.254\ mm

3 0
3 years ago
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