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Mnenie [13.5K]
4 years ago
13

Which simple machine is best used to split apart an object?

Physics
2 answers:
Blizzard [7]4 years ago
4 0
A wedge because you can maneuver it to split something open
mel-nik [20]4 years ago
3 0
The answer would be a wedge (axe).
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What is the gravity force between two stars with mass of 5,000,000 kg and 1,000,000 kg if the distance between them is 100 m
Luda [366]

Answer:

0.0334N

Explanation:

Given parameters:

M1  = 5 x 10⁶kg

M2  = 1 x 10⁶kg

Distance  = 100m

Unknown:

Gravitational force  = ?

Solution:

To solve this problem, we use the Newton's law of universal gravitation.

     Fg  = \frac{G m1 m2}{r^{2} }  

G is the universal gravitation constant

m is the mass

r is the distance

         Fg  = \frac{6.67 x 10^{-11} x 5 x 10^{6}  x 1 x 10^{6} }{100^{2} }   = 0.0334N

8 0
3 years ago
The electric field everywhere on the surface of a thin, spherical shell of radius 0.800 m is of magnitude 902 N/C and points rad
Lilit [14]

Explanation:

(a) From E=

r

2

k

e

​

Q

​

Q=

k

e

​

Er

2

​

=

(8.99×10

9

N⋅m

2

/C

2

)

(8.90×10

2

N/C)(0.750m)

2

​

=5.57×10

−8

C

But Q is negative since  

E

→

 points inward, so

Q=−5.57×10

−8

C=−55.7nC

(b) The negatve charge has a spherically symmetric charge distribution, concentric with the spherical shell

8 0
3 years ago
Long Text (essay)
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An entrepreneur would like their customers to come back, with this the customer service must be good. The owner’s workers must be kind & fair to the consumers. However good your product or service is, the simple truth is that no-one will buy it if they don't want it or believe they don't need it. And you won't persuade anyone that they want or need to buy what you're offering unless you clearly understand what it is your customers really want.

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3 0
3 years ago
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Pls help me i have to submit this due tomorrow :'D
aniked [119]

\textsf {a) As there is no change in the initial and final positions, the displacement is}\\\textsf {equal to 0.}

b) Finding total distance :

Distance travelled from 0 s to 5 s :

  • 1/2 x 5 x 30 = 75 m

Distance travelled from 5 s to 10 s :

  • 5 x 30 = 150 m

Distance travelled from 10 s to 15 s :

  • 150 + 1/2 x 5 x 5
  • 150 + 12.5 = 162.5 m

Distance travelled from 15 s to 20 s :

  • 36 x 5 + 1/2 x 5 x 4
  • 180 + 10 = 190 m

Distance travelled from 20 s to 25 s :

  • 45 x 5 + 1/2 x 5 x 5
  • 225 + 12.5 = 237.5 m

Distance travelled from 25 s to 30 s :

  • 40 x 5 + 1/2 x 10 x 5
  • 200 + 25 = 225 m

Distance travelled from 30 s to 35 s :

  • 20 x 5 + 1/2 x 20 x 5
  • 100 + 50 = 150 m

Distance travelled from 35 s to 40 s :

  • 1/2 x 20 x 5
  • 50 m

Total = 75 + 150 + 162.5 + 190 + 237.5 + 225 + 150 + 50

Total = 1240 m

c) velocity at t = 15 s

  • 36/15
  • 12/5
  • 2.4 m/s

d) average velocity

  • 0 m/s (as displacement is equal to 0)

e) average speed

  • 1240/40
  • 31 m/s

f) Part d uses displacement whereas part e uses distance

3 0
2 years ago
1. What are examples of muscle endurance activities/exercises? 2. Which sports require the most muscle endurance? 3. How many si
Komok [63]

Answer:

hello i need points i need to pass 8th grade

Explanation:

4 0
3 years ago
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