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3 years ago

Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?

2 answers:

8 0

Newton’s Second Law concerns the generation of force based on an object’s mass and acceleration, as described by the equation F=ma.

Hope this helps!

Ronch [10]3 years ago

5 0

The two variables which influences the body force addressed in “Newton’s second law of motion” s mass and acceleration.

<u>Explanation:</u>

The “Newton’s second law of motion” states that rate of change of momentum is directly in proportional to force applied on body.

$\begin{array}{l}{F \propto \frac{d p}{d t}} \\ {F=m \cdot \frac{d v}{d t}} \\ {F=m \cdot a} \\ {\text { Force }=\text {Mass} \times \text {Acceleration}}\end{array}$

This is how the Second law of motion is derived in formula by the Newton which establishes the relationship between the two variables which influences the force of the body.

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The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

3 years ago

I will mark as the brainliest answer<br><br>plz 8,9,10

blagie [28]

Answer:

8. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 unit.

10. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units.

Explanation:

8. i) acceleration = velocity / time

ii) In this figure velocity = time

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. i) acceleration = velocity / time

ii) In this figure 4 = m + 5, therefore m = -1

therefore velocity = (-0.5 $\times$ time) + 5

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 units.

10.) velocity is constant at 2

therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units

5 0

3 years ago

Amy counts the wave crests traveling down a stretched string.Five wave crests pass Amy in two seconds.What is the frequency of t

Verdich [7]

The frequency of a wave is a measure of the number of waves that passes through a point per unit of time. It has SI units of s^-1. It is also equivalent to Hertz (Hz). We calculate the frequency of the wave described as follows:

Frequency = 5 waves / 2 s = 2.5 / s or 2.5 Hz

4 0

2 years ago

The answer can not be C or D its between A and B

Murrr4er [49]

If the answer is not C D then is A bc of the way it crosses over

8 0

3 years ago

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to

DerKrebs [107]

Answer:

A) 26.5 m/s

B) 33.0 m/s

Explanation:

Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.
So, we can apply the definition of average velocity to find this speed as follows:

$v_{x} = \frac{\Delta x}{\Delta t} (1)$

We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.
This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.
Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).
Since the acceleration is constant, we can use the following kinematic equation, as follows:

$\Delta y = y_{f} - y_{o} = v_{o} * t + \frac{1}{2} * g *t^{2} (2)$

if we take the river level as our x-axis, this means that yf = 1.3 m and

y₀ = 20.8 m.

At the same time, due to in the vertical direction the car has no initial velocity, this means that v₀ = 0.
Replacing by the values in (2) , and solving for t:

$t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} } = 2 s (3)$

If we choose t₀ =0 ⇒ Δt = t = 2 s
Replacing Δx and Δt in (1):

$v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s (4)$

When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.
Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:

$v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s (5)$

Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:

$v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}} = 33.0 m/s (6)$

5 0

2 years ago