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mixer [17]
3 years ago
15

A force of 8 N accelerates by 4 m/s^2. What would be the amount of force needed to give a final acceleration of 5.3 m/s^2

Physics
1 answer:
Elis [28]3 years ago
4 0

Answer:

10.6 N

Explanation:

F = ma

8 = m * 4

m = 2 kg

F = 2 * 5.3

F = 10.6 N

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Precipitation is greater over what and less over what?<br><br> A. Land, sea<br><br> B. Sea, land
Ksju [112]
I would say letter A to this because we need rain water for crops plants. Also we need it so that we can drink it. Its less great over sea because as it merge with the salt in the ocean it becomes less drinkable and less usable.
3 0
3 years ago
The bellow of a territorial bull hippopotamus is measured at 100 dB above the threshold of hearing. What is the sound intensity?
Dmitriy789 [7]

Answer:

Explanation:

The hippopotamus hearing threshold is 100dB

β = 100 dB

The threshold of human hearing is Io = 1 × 10^-12 W/m²

The sound intensity level is given as

β = 10•Log(I / Io)

100 = 10•Log(I / Io)

Divided Both sides by 10

100 / 10 = Log(I / Io)

10 = Log(I / Io)

Take inverse Logarithm ( antilog) of both sides

10^10 = 10^[Log(I / Io)]

10^10 = I / Io

Then,

I = 10^10 × Io

I = 10^10 × 1 × 10^-12

I = 1 × 10^-2 W/m²

I = 0.01 W/m²

The sound intensity is 0.01 W/m²

5 0
3 years ago
a pelican flying along a horizontal path drops a fish from a height of 5.4m. the fish travels 8.0m horizontally before it hits t
oksian1 [2.3K]

Answer:

7.0 m/s

Explanation:

6 0
3 years ago
The distance of Saturn from the sun is:<br><br> &lt; 1 A.U.<br> &gt; 1 A.U.<br> = 1 A.U.
sertanlavr [38]

Answer:

1 astronomical unit is the average distance from the Earth to the Sun; approximately 150 million km. At its closest point, Saturn is 9 AU, and then at its most distant point, it's 10.1 AU. Saturn's average distance from the Sun is 9.6 AU. We have written many articles about Saturn for Universe Today.

Explanation:

6 0
3 years ago
Read 2 more answers
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

#SPJ4

7 0
2 years ago
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