New ocean crust or sea floor is produced when tectonic plates ______

If the child pulls with a force of 20 N for 12.0 m , and the handle of the wagon is inclined at an angle of 25 ∘ above the horiz

lilavasa [31]

Answer: 217.52 N

Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.

Hence the formulae that defines work done is given by

W = Force × distance

But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.

Hence

Work done = (horizontal component of force) × distance

Work done = F cos θ × distance

Work done = 20 cos 25 × 12 = 217.52 N

4 0

3 years ago

Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it

d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s, 1) v = $\sqrt{0.55^2 + 19.6 y}$ , 2) v = 2.05 m / s,

3) d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

Q = v A

where Q is the flow rate, A is area and v is the velocity

the area of a circle is

A = π r²

radius and diameter are related

r = d / 2

substituting

A = π d²/4

Q = π/4 v d²

let's reduce the magnitudes

v = 0.55 m / s = 55 cm / s

let's calculate

Q = π/4 55 1.96²

Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

v = $\sqrt{0.55^2 + 2 \ 9.8\ y}$

v = $\sqrt{0.55^2 + 19.6 y}$

2) ask to calculate the velocity for y = 0.2 m

v = $\sqrt{0.55^2 + 19.6 \ 0.2}$

v = 2.05 m / s

3) We write the continuous equation for this point 2

Q = v₂ A₂

A₂ = Q / v₂

let us reduce to the same units of the SI system

Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

A₂ = 165.95 10⁻⁶ / 2.05

A₂ = 80,759 10⁻⁶ m²

area is

A₂ = π/4 d₂²

d₂ = $\sqrt{4 A_2 / \pi }$

d₂ = $\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }$

d₂ = 10.14 10⁻³ m

d₂ = 1.014 cm

4 0

3 years ago

You are writing a report about space travel. Where would be the best place to find information? (2 points)

Harrizon [31]

Answer: Contact the National Aeronautics and Space Association

Explanation:

The National Aeronautics and Space Association is an independent organization in US which aims to enable a safe, secure, efficient space journey and launch of new satellites, spacecraft beyond the earth's orbit. It can give better information for the space travel already conducted by the astronauts. Thus will help in writing the report as it is the best place to find the information.

5 0

3 years ago

Read 2 more answers

Swinging a tennis racket against a ball is an example of a third class lever. Please select the best answer from the choices pro

kari74 [83]

that statement is true

a Third class lever applied when the effort place between the load and the fulcrum.

For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm

7 0

3 years ago

Read 2 more answers

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

New ocean crust or sea floor is produced when tectonic plates _______. diverge converge