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Sergeeva-Olga [200]
3 years ago
13

Pls help asap

Physics
2 answers:
RoseWind [281]3 years ago
8 0
I’m pretty sure the answer is a
Ira Lisetskai [31]3 years ago
3 0

Answer:

D

Explanation:

i took lots of time on the lesson

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Do you get this ?????
Jet001 [13]

The first one is actually 10 times as big as the second one.

Because of their places, the first one means 6000, and the second one means 600.

8 0
3 years ago
98 POINTS, 5 simple questions!! HELP
lozanna [386]

25,000 Feet = 7620m

PE = mgh where m is mass, g is gravity accel: 9.8 n h is height

= 90 x 9.8 x 7620

= 6720840J

= 6.72MJ

F = ma where m is mass, a is accel = gravity = 9.8

= 90 x 9.8

= 882N

Accel = gravity = 9.8m/s^2

KE = 1/2mv^2 where m is mass n v is vel

if no wind resistance, PE leaving airplane = KE at net

6720840 = 1/2 x 90 x v^2

v^2 = 149352

v = 386.5m/s


3 0
3 years ago
Read 2 more answers
An object with a mass of 10 kg is accelerated upward at 2 m/sec2. What force is required?
telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

6 0
3 years ago
Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete
harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

v=\dfrac{d}{T}

d is distance, d=2\pi r and 1/T = f (frequency)

v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s

So, the average speed of the rubber stopper is 18.84 m/s.  

4 0
3 years ago
ich of the following is not accurate when describing solids? A. The amount of pressure exerted by a solid is solely dependent on
Dafna11 [192]
The amount of solid does not affect how you are describing the solid so a is the answer
5 0
3 years ago
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