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siniylev [52]
3 years ago
14

What is the relationship between the perimeter and the side length of anequilateral triangle ?

Physics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90. Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

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In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times
Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0
3 years ago
Name the types of forces​
Lostsunrise [7]
Air resistance force
tension force
spring force
frictional force
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6 0
3 years ago
Benjamina started her walk from the front door of her ground floor apartment. She walked 6 meters to the corner of the building
Cloud [144]
The distance is the total distance she walked which is 16 meters adding the 6 meters to the corner and 10 meters to her friend's apartment. Her displacement is the distance from her original starting point so you set up a triangle with side lengths of 6 and 10 and solve for the hypotenuse which gives you a displacement of 11.66 meters.  
5 0
3 years ago
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True or False: The basketball should be dribbled below the waist.
zimovet [89]
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.
7 0
2 years ago
When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi
Nikitich [7]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} .

Comparing two wires,

R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}

Dividing the above 2 equations,

\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{A_2 }{A_1}  \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{d_2^2 }{d_1^2}  \\

Since d_2=2d_1

The above ratio is

\frac{R_1}{R_2}=\frac{1.68(10^{-8})  }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264

We also have,

\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0
3 years ago
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