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jekas [21]
2 years ago
12

If the applied force (F) is 14 N and the frictional force (f) is 3 N, what would the acceleration be on this 3.5 kg box?

Physics
1 answer:
Helga [31]2 years ago
4 0

Answer:a = F/m so

11/3.5 times= 3.14 m/s

Explanation:

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Any fracture or system of fractures along which Earth moves is known as a D.fault.

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3 0
3 years ago
A SPARK PLUG IN AN AUTOMOBILE ENGINE CONSISTS OF TWO
kirill [66]

Answer:

Electric potential, V=3.52\times 10^{12}\ volts

Explanation:

It is given that,

The magnitude of electric field, E=4.7\times 10^7\ V/m

Distance between automobile engines that consists of metal conductors, d = 0.75\ mm = 7.5\times 10^4\ m

Let V is the magnitude of the potential difference between the conductors. The relation between the electric field and the electric potential is given by:

V=E\times d

V=4.7\times 10^7\ V/m \times 7.5\times 10^4\ m

V=3.52\times 10^{12}\ volts

So, the magnitude of the potential difference between the conductors is 3.52\times 10^{12}\ volts. Hence, this is the required solution.

6 0
4 years ago
Friction always opposes an objects?<br><br> A) Power<br> B) Weight<br> C) Motion<br> D) Net force
marin [14]
Friction always opposes motion
8 0
4 years ago
Read 2 more answers
Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​
Mkey [24]

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F  = (1200)(0) = 0

4 0
3 years ago
A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite
DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx  is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x  4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x  4 π

≥  . 5254 x ⁻²⁵

h / λ ≥  . 5254 x ⁻²⁵

 6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ  

12.56 x 10⁻⁹ ≥ λ  

longest wave length = 12.56 n m

6 0
3 years ago
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