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2 years ago

Plz and ty ... free points, be nice

Physics

2 answers:

kiruha [24]2 years ago

7 0

Answer:

oml thx bruh im doing the thing to ummm get 50 rn lol

Explanation:

harkovskaia [24]2 years ago

5 0

Answer:

Thank you!

Explanation:

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The mass of 25cm of ivory was found to be 0.045kg. caculate the density of ivory in SI units

solmaris [256]

Answer:

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Explanation:

,ajsgfgsdfubcygduhdsbcdshyudbckjdshfyendhdndbhfdbuehundweugdjnbfcjdfhewjdnbewqwioehsajcnbsfhhgjhbgjhhbjdhgwbvdwnjjguhbwqhjbsjwqhshwbdjagdasndhsajuwqbdjasnjashdjaskxnjgjjsklx ms,dhwshjksnjdbhjssnj,

7 0

2 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

1 year ago

Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

lana [24]

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

8 0

2 years ago

One of the longer sides of a kite is 39 feet. one of the shorter sides is 25 feet. half the length of the shorter diagonal is 15

pychu [463]

The area of a triangle is found by multiplying the height of the triangle by the length of the base and dividing them both by 2. The length of the shorter side in the equation is useless information, so just multiply 39 by 25 and divide that by 2. A=487.5 sq ft. Also, that's a pretty big kite.

8 0

2 years ago

675 km equals how much cm ?

irga5000 [103]

KHDMDCM.
Now go from Kilometer to Centimeter: 5.
Move the decimal 5 places to the right: 67,500,000 centimeters.
Hope this helps :)

7 0

3 years ago