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vaieri [72.5K]
3 years ago
12

A special triangular frame is being made for a piece of artwork. The the base of the triangular frame must be 90 cm. If the area

of the triangular frame is 765 square cm, what is the height? a) Let h = the height of the triangle. Write the equation you would use to solve this problem.
Physics
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

Explanation:

base of triangular frame, b = 90 cm

Area, A = 765 cm²

Let the height is h.

Area of a triangular frame = 1/2 x base x height

765 = 0.5 x 90 x h

h = 17 cm

Thus, the height of triangular frame is 17 cm.

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Which type of wave interaction is shown in the photo?
julia-pushkina [17]

The wave interaction that is shown in the photo is refraction as light moves from air to water.

<h3>What is refraction?</h3>

Refraction refers to the change in the frequency of a wave and the direction of the wave as it moves from one medium to another. We know that waves makes a body under water to look slightly different than when it is in air.

Thus, the wave interaction that is shown in the photo is refraction as light moves from air to water.

Learn more about refraction:brainly.com/question/14760207

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2 years ago
To be Answered in Sentences...
Nikolay [14]

Answer:

1. The equivalent resistance for the combination of resistors in series is equal to the algebraic sum of all its individual resistances.

2. The Current will increase and causes it to have less restriction.

Explanation:

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8 0
3 years ago
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how difficult is it to start a heavy lorry moving and to stop it moving? (choose one answer from the image provided)
ale4655 [162]

Answer:

option B

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3 0
3 years ago
The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m=11,100 kg is the mass of the telescope

M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
Please someone helppp me.. I will mark as brainliest.​
Mademuasel [1]

Answer:

b) G.P.E = Mgh

300j = M x 10 m/s² x 15 m

300 j/ 10 m/s² x 15 m = M

300j/ 150 s² = M

2kg = M

c) K.E = 1/2 m v²

K.E = 1/2 (50) (50)²

K.E = 1/2 (50) (2500)

K.E= 125000/2

K.E = 625 000 J

4 0
3 years ago
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