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jarptica [38.1K]
3 years ago
6

Which two terms describe the relationship between the two variables in this graph?

Physics
1 answer:
harkovskaia [24]3 years ago
6 0
Show a picture of the graph
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Which of the following processes is not a physical change:drying wet clothes,cutting snowflakes out paper,lighting a match from
gavmur [86]
Cutting snowflakes out of a paper
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3 years ago
Yeah, so, my teacher finna kill me if I don't turn this in today..
AnnZ [28]

Answer:

Blank 1: 10

Blank 2: 4

Blank 3: 10

Blank 4: 5

Blank 5: 5

7 0
3 years ago
The game was invented in Holyoke, Mass<br> True or false?
zheka24 [161]

Answer:

true

Explanation:

Throwback Thursday: When Volleyball Was Invented in Massachusetts. On this day in 1895, a Holyoke man created a game called 'mintonette

6 0
3 years ago
Suppose a force of 30 N is required to stretch and hold a spring 0.2 m from its equilibrium position.
nydimaria [60]

Answer:

a. k=150N/m, b.W=0.75J,c. W=12J, d.W=9J

Explanation:

a. The Hooke's Law statesF=kx, where F is the force, k the spring constant and x the displacement, Knowing the force and the displacement you can find k:

F=kx

k=\frac{F}x=\frac{30}{0.2}=150N/m

b. The work done by a force F(s) that moves along a displacement s is:

W=\int\limits^{x_2}_{x_1} {F(s)} \, ds

ThenF(s)=ks (Hooke's Law)

W=\int\limits^{x_2}_{x_1} {F(s)} \, ds=\int\limits^{x_2}_{x_1} {ks} \, ds=\frac{1}{2} ks^2|\limits^{x_2}_{x_1}=\frac{1}{2} k(x_2^2-x_1^2)

Work needed to go from  x_1=0 to x_2=0.1

W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.1^2-0^2)=75(0.01)=0.75 J

c. Work needed to go from x_1=0 to x_2=0.4

W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.4^2-0^2)=75(0.16)=12 J

d. Work needed to go from x_1=0.2 to x_2=0.4

W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.4^2-0.2^2)=75(0.16-0.04)=75(0.12)=9 J

5 0
3 years ago
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x
mezya [45]

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

3 0
3 years ago
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