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-BARSIC- [3]
2 years ago
5

Pls qmp

Physics
1 answer:
AVprozaik [17]2 years ago
8 0

Answer: I put the importance of the lab in the topic is to find how dense an object is and if it can sink or float and it's important to answer the question so you can also find mass and volume.

Explanation: I dunno haha...

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A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d
lesya [120]

Answer:

d_{2}=0.585m

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

T_{1}-T_{2}=0

T_{1}=T_{2}

F_{1}*d_{1}=F_{2}*d_{2}

43N*0.30m=21N*d_{2}

Solve to d2

d_{2}=\frac{41N*0.30m}{21N}

d_{2}=0.585m

5 0
3 years ago
You ride your skateboard to school and tell all your friends you rode at speed of 25 miles per hour
Ray Of Light [21]
And they think you're cool
5 0
3 years ago
Fluids can flow and change shape to fit their containers. Which two states are<br> fluids?
mr_godi [17]

Answer: the two states that are fluid are;-

<u>#{1} liquid</u>

<u>#{2} gas </u>

Explanation:

as we know that there are mainly three states of substance

but among them only two of them can fluid and takes the shape of the container that are liquid and gas

4 0
2 years ago
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr
stealth61 [152]

Answer:

\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

R=\frac{\rho L}{A}

\rho is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}

For the third and fourth coils, we have:

R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

6 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
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