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3 years ago

How is the acceleration of an object in newtons second law related to its mass and the net force on it?

Physics

2 answers:

ki77a [65]3 years ago

8 0

Answer:

C) Acceleration is directly proportional to net force divided by mass.

Explanation:

As per Newton's 2nd law we know that net force on the object is product of mass and acceleration of the object

so we will have

F = ma

so here we can say that the force and mass of the object is directly depends on each other and the equation is just product of mass and acceleration.

So in order to find the acceleration we have

$a = \frac{F}{m}$

so we will have

C) Acceleration is directly proportional to net force divided by mass.

9966 [12]3 years ago

4 0

B) Acceleration is directly proportional to the mass of the object

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7. In a football game, the running back is running up the field. He starts from rest and runs for 3 seconds

maria [59]

Answer:

11.25m

Explanation:

Given parameters:

Initial velocity = 0m/s

Time of running = 3s

Acceleration = 2.5m/s²

Unknown:

Displacement = ?

Solution:

To solve this problem, we apply one of the motion equations.

S = ut + $\frac{1}{2}$ at²

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t is the time taken

a is the acceleration

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2 years ago

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

Mariana [72]

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

Total Energy before fall = Total Energy after fall

$E_{i}$ = $E_{f}$

Here, total energy is the sum of kinetic energy and potential energy

$K.E_{i}$ + $P.E_{i}$ = $K.E_{f}$ + $P.E_{f}$ (a)

When banana is at height h, it has

$K.E_{i}$ = 0 and $P.E_{i}$ = mgh

and when it reaches the river, it has

$K.E_{f}$ = 1/2m $v^{2}$ and $P.E_{f}$ = 0

Putting the values in equation (a)

0 + mgh = 1/2m $v^{2}$ + 0

mgh = 1/2m $v^{2}$

cutting 'm' from both sides

gh = 1/2 $v^{2}$

v = $\sqrt{2gh}$

Hence, the velocity of banana before hitting the water is

v = $\sqrt{2gh}$

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3 years ago

At a certain instant, the charge stored by a capacitor in an LC circuit is 60 C and the energy stored by the capacitor is nine

rusak2 [61]

Complete Question: At a certain instant, the charge stored by a capacitor in an LC circuit is 60 microC and the energy stored by the capacitor is nine sixteenths of its maximum value. If the circuit's frequency of oscillation is (40 /pi) Hz, what is the maximum current in the circuit?

Answer:

Imax = 6.4 mA

Explanation:

In a oscillating LC circuit, energy is exchanged all time between the capacitor and the inductor, in such a way that any time, the total energy in the circuit is equal to the sum of the energies stored in both circuit elements.

When the capacitor is fully discharged, the energy is stored only at the inductor, so the current is maximum.

We can write the following equation in this case:

1/2 L*Imax² = 1/2 Li² + 1/2q₀²/C

We have as givens, q₀= 60μC and f=40/π Hz.

We also know, that when the capacitor is charged with this q₀, the energy stored in it (actually in the electric field between plates) is 9/16 of its maximum value, which must be equal to the maximum energy stored in the inductor (due to the energy conservation), so, we can write the following equation:

1/2 q₀²/C = 9/16*1/2*L*Imax²

In a pure oscillating circuit, there is a fixed relationship between f, L and C, as follows:

ω₀² = 1/LC ⇒ LC = 1/ω₀²

Simplifying common terms, and manipulating terms, we have:

I²max =16*q₀²*(2*π*f₀)² = 16* (60)²*4*π²*((40)/π)² / 9

⇒ I²max = 16*(60)²*(40)²*4*10⁻¹²/9 A

⇒ I max = ((8*60*40) / 3) * 10⁻⁶ A = 6.4*10⁻³ A = 6.4 mA

3 0

3 years ago