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frez [133]
2 years ago
10

A block is pulled across a table by a constant force of 9.20 N. If the mass of the block is 2.30kg, how fast will the block be m

oving after 2.00 seconds? Assume negligible friction.
Physics
1 answer:
kaheart [24]2 years ago
5 0

Answer:

8\:\text{m/s}

Explanation:

From Newton's 2nd Law, we have \Sigma F=ma. Using this, we can find the acceleration of the object:

9.20=2.30a,\\a=\frac{9.20}{2.30}=4\:\mathrm{m/s^2}.

Now that we've found the block's acceleration, we can use the following kinematics equation to find its final velocity after 2 seconds:

v_f=v_i+at,\\v_f=0+4(2),\\v_f=\boxed{8\:\text{m/s}}

*Assumption: The block is initially at rest and has a initial velocity of zero. Otherwise, the question is unsolvable.

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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

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3 years ago
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Tema [17]

Answer:

D

Explanation:

The answer is Niels Bohr's planetary model, the difference between this model and all of the other models is that the Bohr's PM Is more of layers of

Nucleus - Protons and Neutrons

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Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters
Vesnalui [34]

Answer:

C. \frac{3F}{8}

Explanation:

Let initial charges on both spheres be,q

F=\frac{Kq^2}{d^2}   \ \ \  \ \ \  \ \ \  \ \_i

When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

#Now, when C is touched by B, the final charges on both of them will be:

q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

Now the force between A and B is calculated as:

F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

Hence the electrostatic force becomes 3F/8

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