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hjlf
3 years ago
12

The umbra during a SOLAR OR LUNAR eclipse is smaller than during a SOLAR OR LUNAR eclipse.

Physics
1 answer:
skad [1K]3 years ago
3 0
The umbra during a solar eclipse is smaller than during a lunar eclipse.
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In object of mass 30 kg is in freefall in a vacuum where there is no air resistance determine the acceleration of the object
Leto [7]
The free fall acceleration due to the Earth's gravity will always be 9.8m/s^2. So, if this question is taking place on Earth, then the acceleration of an object in free fall is 9.8m/^2.
3 0
3 years ago
3. Why is the term cold blooded a misconception? Explain​
jok3333 [9.3K]

Answer:

The term “cold-blooded” implies that these animals are in a never-ending struggle to stay warm. That really isn't correct. A cold-blooded animal can warm up their blood by being in the sun for hours.

7 0
3 years ago
The pressure of a box pushes down on the floor is 50 Pa if the box weighs 400 N what is the area of the base of the box
Wewaii [24]

Answer:8m^2

Explanation:

Area=force÷pressure

Area=400÷50

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5 0
3 years ago
a 1.2 m wire carries a current of 10.0 A in a uniform magnetic field of 0.050T. Find the magnitude of the magnetic force on the
rodikova [14]
To calculate we use the formula for a magnetic force in a current-carrying wire expressed as the product of the current, magnetic field and the length of the wire.
 F = I x L x B

where F is the force on the wire, I is the current flowing on the wire, L is the length of the wire and B is the magnetic field.

F = 10.0 A x 1.2 m x 0.050 T
F = 0.60 N
8 0
3 years ago
An object is placed 10 cm in front of a diverging mirror. What is the focal length of the mirror if the image appears 2 cm behin
Dafna1 [17]

Answer:

the focal length of the mirror is :  f=-2.5\,\,cm

Explanation:

Use the formula for the formation of image using a divergent mirror and recalling that the image (s') that this mirror formed is virtual, so it is entered as a negative number in the formula. Use the object position (s) as 10, the image position (s') as -2, and derive the value of the focal length:

\frac{1}{s} +\frac{1}{s'}=\frac{1}{f}\\\frac{1}{10} +\frac{1}{-2}=\frac{1}{f}\\\frac{1}{10} -\frac{1}{2}=\frac{1}{f}\\\frac{10\,f}{10} -\frac{10\,f}{2}=\frac{10\,f}{f}\\f-5\,f=10\\-4\,f=10\\f=-2.5\,\,cm

6 0
3 years ago
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