What power of an engine is required to pump 2450N of water per second from a well 50m deep to the surface

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0

3 years ago

A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(

Flura [38]

Answer:

8.97 Watt

Explanation:

Resistance, R = 20 ohm

Inductance, L = 10 mH

V(t) = 20 Cos (1000 t + 45°)

Compare with the standard equation

V(t) = Vo Cos(ωt + Ф)

Ф = 45°

ω = 1000 rad/s

Vo = 20 V

Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm

impedance is Z.

$Z = \sqrt{R^{2}+X_{L}^{2}}$

$Z = \sqrt{20^{2}+10^{2}}$

Z = 22.36 ohm

$V_{rms}=\frac{V_{0}}{\sqrt{2}}$

$V_{rms}=\frac{20}{\sqrt{2}} = 14.144 V$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{14.144}{\sqrt{22.36}}=0.634 A$

Apparent power is given by

P = Vrms x Irms

P = 14.144 x 0.634

P = 8.97 Watt

6 0

3 years ago

Earth can be modeled with five layers based on the strengths and physical properties of the materials in them. Put the layers in

andre [41]

Answer:

Crust, Upper mantle, mantle, outer core, inner core

Explanation:

The Earth's layers have been clasified in 5 according to the materials that conform them, theri physical properties, strengths and also their state of matter. We all know how the outer layer of the Earth looks like, but if we start to dig a huge hole we are going to see different types por materials due to a change in pressure, temperature, and other factors. At the very center of the Earth there's what's called "core". The core is liquid and at extremely high temperatures. This is because of the enormous amount of pressure the rest of the Earth is putting it under. So, if we list the different layers of the Earth according to the materials they are made of, from the Earth's surface to the core, the answer is:

1) Crust (surface)

2) Upper Mantle

3) Mantle

4) Outer core

5) Inner core

In some books you may find a layer called Lithosphere. Tis layer consists not only of the crust, but also it contains the transition zone between the upper mantle and the crust.

4 0

3 years ago

A 143 kg astronaut is floating in

Dima020 [189]

The recoil velocity of the astronaut is -0.070 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the astronaut-wrench system must be conserved.

At the beginning, their total momentum is zero:

$p=0$ (1)

Later, after the astronaut throws the wrench, the total momentum is

$p=mv+MV$ (2)

where

m = 0.725 kg is the mass of the wrench

v = 13.8 m/s is the velocity of the wrench

M = 143 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since momentum is conserved, (1) = (2), and so we can find V:

$0=mv+MV\\V=-\frac{mv}{M}=-\frac{(0.725)(13.8)}{143}=-0.070 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

5 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

3 years ago