If a data point is way off the trend line which of the following will not help resolve the problem

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

7 0

3 years ago

Question 2 of 10

Mekhanik [1.2K]

the awnser to ur question is D

6 0

3 years ago

A CAR ACCElerates FROM REST To

Ilia_Sergeevich [38]

Answer:

666.6 kW

Explanation:

Power Formula

Power = Force × Velocity

Calculating Force

F = ma
F = m(v - u / t) [From the equation v = u + at]
F = 2,000 (50/7.5)
F = 2,000 (20/3)
F = 40,000/3
F = 13333.3 N

Solving for Power

P = 13333.3 × 50
P = 666666.6 W
P = 666.6 kW

5 0

1 year ago

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago