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Mashcka [7]
3 years ago
5

If javier has a car thats 250,000 kg and a forse of 300 newtons.what would be the accleration.

Physics
1 answer:
const2013 [10]3 years ago
3 0

Answer:

The acceleration would be 3.455.

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Compare and contrast series and parallel circuits?
dalvyx [7]

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

Hope It Helps!

6 0
3 years ago
6. A skier starts from rest at the top of a frictionless incline of height 20.0 m. At
n200080 [17]

Explanation:

a. KE at bottom = PE at top

½ mv² = mgh

v = √(2gh)

v = √(2 × 9.8 m/s² × 20.0 m)

v = 19.8 m/s

b. Work by friction = PE at top

mgμ d = mgh

d = h / μ

d = 20.0 m / 0.210

d = 95.2 m

6 0
3 years ago
Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, stick together and move off after
Natali [406]

Answer:C The final velocity is half of trained car B's initial velocity.

Explanation:

7 0
3 years ago
Read 2 more answers
Please help, if u can. i linked the picture down below :)
natali 33 [55]

Answer:

For Parent #1 missing box it's t and for Parent #2 missing box is t

5 0
3 years ago
A block is sliding down an inclined plane that makes an angle of 65o with the horizontal. If the coefficient of kinetic friction
Nezavi [6.7K]

Answer:

8.2 m/s²

Explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17

F_{n} = Normal force on the block by the ramp

f_{k} = kinetic frictional force

Force equation perpendicular to ramp surface is given as

F_{n} = mg Cos65

Kinetic frictional force is given as

f_{k} = \mu F_{n}

f_{k} = \mu mg Cos65

Force equation parallel to ramp surface is given as

mg Sin65 - f_{k} = ma

mg Sin65 - \mu mg Cos65 = ma

g Sin65 - \mu g Cos65 = a

(9.8) Sin65 - (0.17) (9.8) Cos65 = a

a = 8.2 m/s²

4 0
3 years ago
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