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DIA [1.3K]
2 years ago
10

Electromagnetic wav​

Physics
2 answers:
krek1111 [17]2 years ago
6 0

Answer:

I dont get what your asking sorry boo

Explanation:

Sliva [168]2 years ago
3 0

Answer:

e search mo sa utak mo kasi di mo alam

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How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?
soldi70 [24.7K]

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

6 0
3 years ago
Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th
tatiyna

Answer:

b) -10 m/s

Explanation:

In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.

3 0
2 years ago
A student claims that the Chernobyl accident proves that nuclear power is on safe for use in the United States which statement i
Setler79 [48]
Hi! Check out my valid counter argument below!

"The accident only released harmless gamma rays."

Hope I helped!
5 0
3 years ago
41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h
romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0
3 years ago
A glass of juice at 20°C was kept in the open. After 30 minutes, the temperature of the Juice remained the same. Which statement
devlian [24]

The answer would be B the temperature of the juice was the temperature of the surrounding air

7 0
3 years ago
Read 2 more answers
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