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bija089 [108]
3 years ago
6

How much force is required to accelerate a 50 kg mass at 2 m/s2?

Physics
1 answer:
skelet666 [1.2K]3 years ago
5 0
Newton's 2nd law of motion: 

                    Force = (mass) x (acceleration)

                              = (50 kg) x (2 m/s²)

                              =    100 newtons .                                   
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A symbolic model for learning is a model that is observed in person.
kap26 [50]

Answer:

True

Explanation:

3 0
3 years ago
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The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st
RSB [31]

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

The car travels approximately 17.43m before it stops

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6 0
2 years ago
In which situation will the lowest resistance occur?
insens350 [35]

Answer:

thick wire and cold temperatures

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3 years ago
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66 POINTS! Which is true about the four atoms shown in figures A, B, C, and D?
Butoxors [25]

Answer:

A

Explanation:

A and B are isotopes of one another but the same element

C and D are isotopes of one another but the same element

However, A and B have a different proton count than C and D, indicating different elements because the proton count is equivalent to the atomic number.

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2 years ago
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A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

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