A vector is a quantity that has what two items

(answer it or get reported) Please answer it w the steps

Travka [436]

Answer;

The mass value for the above kinetic energy equation is 400.0000 kg. This is equal to:

■ 400,000.0000 g.

■ 14,109.6000 ounces.

■ 881.8480 pounds.

3 0

3 years ago

At which type of boundary between two tectonic plates is an earthquake or a tsunami most likely to form?

Fofino [41]

Answer:

B. convergent boundary

Explanation:

Convergent boundaries, where ocean plates are subducted at trenches, produce the most tsunamis because they produce the largest earthquakes

7 0

3 years ago

A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical

svp [43]

Answer:

X-Positions: Y-Positions

x(0) = 0 y(0) = 0

x(2) = 120 m y(2) = 19.6 m

x(4) = 240 m y(4) = 78.4 m

x(6) = 360 m y(6) = 176.4 m

x(8) = 480 m y(8) = 313 m

x(10) = 600m y (10) = 490 m

Explanation:

X-Positions

First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

$x = v_{ox} * t = 60.0 m/s * t(s)$

It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
At any moment, it is subject to the acceleration of gravity, g.
As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

$y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}$

Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
y(2) = 2* 9.8 m/s2 = 19.6 m
y(4) = 8* 9.8 m/s2 = 78.4 m
y(6) = 18*9.8 m/s2 = 176.4 m
y(8) = 32*9.8 m/s2 = 313.6 m
y(10)= 50 * 9.8 m/s2 = 490.0 m

5 0

3 years ago

A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

2 years 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