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mylen [45]
3 years ago
12

Creates an image that appears upside down behind the focal point

Physics
1 answer:
zheka24 [161]3 years ago
8 0
An image that appears upside down behind the focal point is an image that is reflected on a concave mirror. Mirrors reflect different kinds of images based on the placement of an object that is reflected towards it. There are two kinds of mirrors, concave and a convex mirrors, the latter makes objects seem smaller and farther than where it is exactly.
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Two rockets are being tested. Both rockets have the same mass. One rocket accelerates at a higher rate than another rocket. What
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Answer:

d

Explanation:

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3 years ago
Why do atoms do group 1 element lose electrons to form cations
Mrac [35]

Answer:

when a element of 1 group take part in reaction, its atom looses outer electron and form positively charged ions called Cation.

Explanation:

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3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

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7 0
2 years ago
For an object that is speeding up at a constant rate how would the acceleration vs. time graph look?
frosja888 [35]

Answer:

C

Explanation:

Horizontal is constant... if it was on x axis it would be a speed of 0

Please give brainliest answer

5 0
3 years ago
Read 2 more answers
A car initially traveling at 27.7 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s2 after its brakes are app
lys-0071 [83]

The car's velocity at time <em>t</em> is given by

v=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t

It comes to a stop when <em>v</em> = 0, which happens when

0=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t\implies t=13.85\,\mathrm s

or after about 13.9 s.

In this time, the car travels a distance <em>x</em> given by

x=\left(27.7\dfrac{\rm m}{\mathrm s}\right)(13.85\,\mathrm s)+\dfrac12\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m

or about 192 m.

In one complete revolution, each tire covers a distance equal to its circumference,

2\pi(0.340\,\mathrm m)\approx2.13628\,\mathrm m

or about 2.14 m.

This means each tire will complete approximately 192/2.14 ≈ 90 revolutions.

8 0
3 years ago
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