An engine is used to lift a beam weighing 9,800 N up to 290 meters.
How much work must be done to lift the beam?

Answer:

B

Explanation:

Answer:

A combination of two or more substances.

Answer:

90 A

Explanation:

Applying,

P = VI.............. Equation 1

Where P = Power, V = Supply voltage, I = current.

Make I the subject of the equation

I = P/V.............. Equation 2

From the question,

Assuming the Voltage of the main electric supply is 120 V

Given: P = 10.8 kW = 10800 W, V = 120 V (household voltage)

Substitute these values into equation 2

I = 10800/120

I = 90 A

Answer:

F = 2.45 N

Explanation:

m = 0.25 kg

a = 9.8 m/s^2

F = ma

F = (0.25)(9.8)

F = 2.45 N

Answer:

Explanation:

The Law of Momentum Conservation applies, and the formula for us here specific to this problem is:

[tex][(m_bv_b)+(m_mv_m)]_b=[(m_b+m_m)v_{both}]_a[/tex] and filling in:

[(1.50*4.50)+60.0*0)] = [(1.50 + 60.0)v] and

6.75 + 0 = 61.5v so

v = .110 m/s in the same direction as the ball

Answer:

The motorcycle daredevil does not sail across the 10-m wide pool filled with hungry crocodiles

Explanation:

The given parameters are;

The height of the ramp = 2.0 m-high

The angle of elevation of the ramp, θ = 20°

The width of the pool = 10 m

The speed with which he is going when the engine fails = 10.4 m/s

The rolling friction between the rubber tires, μ = 0.02

μ·g·cos(θ) = (1/2)·(a)

a = (μ·g·cos(θ))/(1/2)

Where;

a = The acceleration down the plane

∴ a = 0.02 × 9.81 × cos(20°)/(1/2) ≈ 0.369

The length of the ramp, l = 2.0 m/(sin(20°)) ≈ 5.8476088 m

The velocity at the end of the ramp, v² = u² - 2·a·l

∴ v² = 10.4² - 2×(0.369 + 9.81 × sin(20°))×5.8476088 ≈ 64.60446471

v ≈ √(64.60446471) ≈ 8.04

∴ The velocity at the top of the ramp, v ≈ 8.04 m/s

The time it takes the motorcycle to reach ground level from the top of the ramp, t, is given as follows;

h = h₀ + u·t + (1/2)·a·t²

h₀ = 2, u = 8.04×sin(20°), a = -9.81

When h = 0 (ground level), we have;

∴ h = 2+ 8.04×sin(20°)·t - (1/2)×9.81×t²

0 ≈ 2.75·t - 4.905·t² + 2

2.75·t - 4.905·t² + 2 = 0

By using a graphing calculator, we have, t = 0.978 or t = -0.42

At ground level, the time of flight, t ≈ 0.978 s

The horizontal distance the motorcycle travels, before landing, x = u × cos(θ) × t

∴ x = 8.04×cos(20 Degrees) × 0.978 = 7.3889

The horizontal distance the motorcycle travels, before landing, x ≈ 7.3889 m

Therefore, given that the horizontal distance the motorcycle travels in the air 7.3889 m < 10 m, therefore the motorcycle daredevil does not sail across the 10-m wide pool filled with hungry crocodiles

Answer:

[tex]momentum = mass \times velocity \\ = 7.5 \times 1.6 \\ = 12 \: kg {ms}^{ - 1} [/tex]

Mass is 7.5 kg

Velocity is 1.6 m/s

we know that,

momentum = mass × velocity

or; momentum =7. 5 kg × 1. 6 m/s

or; momentum = 12 kgms^1

Answer:

F=Gxy/r^2

Explanation:

F=Gm1m2/d^2

F=Gxy/r^2

Answer:a gas at pressure of 0.8 m column of H g

Answer:

718.02N is the answer to the question

Answer:

[tex]x = 2\sqrt 2[/tex]

Explanation:

Given

[tex]Hypotenuse = 4[/tex]

Required

Find x

Since the triangle is a 45-45-90 triangle, the following relationship exists

[tex]x^2 + x^2 = 4^2[/tex] --- i.e. the other legs are equal

So, we have:

[tex]2x^2 = 16[/tex]

Divide both sides by 2

[tex]x^2=8[/tex]

Take square roots of both sides

[tex]x = \sqrt 8[/tex]

Simplify

[tex]x = 2\sqrt 2[/tex]

Answer:

7√2

Explanation:

Answer:

Power = 1800 W

Explanation:

Power = work / time

Power = 4500 / 2.5

Power = 1800 W

Answer:

[tex]0.5\; \rm m^{3}[/tex].

Explanation:

Calculate the mass of this object:

[tex]\begin{aligned}& m(\text{object}) \\ &= \rho(\text{object}) \cdot V(\text{object}) \\ &= 500\; \rm kg \cdot m^{-3} \times 1\; \rm m^{3} = 500\; \rm kg\end{aligned}[/tex].

Multiple the mass of this object by the gravitational field strength, [tex]g[/tex], to find the weight of this object: [tex]W(\text{object}) = m(\text{object}) \cdot g[/tex].

Since this object is floating in water, the buoyancy force on it should be equal to its weight:

[tex]F(\text{buoyancy}) = W(\text{object}) = m(\text{object}) \cdot g[/tex].

By Archimedes' Principle, the weight of the water that this object displaces would be equal to [tex]F(\text{buoyancy})[/tex], the size of buoyancy force on this object,

Hence, the weight of water displaced would be [tex]W(\text{object}) = m(\text{object}) \cdot g[/tex].

Divide this weight by [tex]g[/tex] to find the mass of water displaced:

[tex]\begin{aligned} & m(\text{water displaced}) \\ &= \frac{W(\text{water displaced})}{g} \\ &= \frac{m(\text{object}) \cdot g}{g} = m(\text{object}) = 500\; \rm kg\end{aligned}[/tex].

Assume that the density of water is [tex]\rho(\text{water}) = 1000\; \rm kg \cdot m^{-3}[/tex]. The volume of water displaced would be:

[tex]\begin{aligned}& V(\text{water displaced}) \\ &= \frac{m(\text{water displaced})}{\rho(\text{water})} \\ &= \frac{500\; \rm kg}{1000\; \rm kg \cdot m^{-3}} = 0.5\; \rm m^{3}\end{aligned}[/tex].

Answer:

3386.23 Km

Explanation:

Without any wind interference the speed of the plane, it flies at 2.42 x 102 m/s or 242 m/s

The magnitude of the velocity of the air = 40.7-m/s

The time taken, for the plane to fly due west, tw = distance, x ÷ (speed of the plane without wind interference, up - the magnitude of the velocity of the air, ua).......... (equation 1)

The time taken, for the plane to fly due east, te = distance, x ÷ (speed of the plane without wind interference, up + the magnitude of the velocity of the air, ua).......... (equation 2)

To calculate the total time, t, the plane can fly to cover east and west distances, we add equation 1 and 2 together

t = (x ÷ (up - ua)) + (x ÷ (up + ua))

Making distance x, the subject of the formula we have:

x = t ( up²- ua²) ÷ 2up

note t = 8.0 hours which is (8 x 60 x 60) seconds = 28,800‬

So, x = 28,800‬ ((242 m/s)² - (40.7-m/s)²)÷ 2(242 m/s)

x= 1,638,936,288‬ ÷ 484

= 3386.23 Km

The power required to do 180 J of work in 2.4 s is 75 W.

Power can be defined as the rate at which work is done.

To calculate the power, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the power required to do 180 J of work in 2.4 s is 75 W.

Learn more about power here: https://brainly.com/question/24858512

#SPJ2

Answer:

Some galaxies are given descriptive names (e.g. "Andromeda", "Whirlpool") if they are particularly distinctive in location or appearance. But most galaxies are known from their designation in a catalogue. One of the earliest catalogues of objects in the sky was made by Charles Messier.

4.9 eV is the correct answer

Hope it helps you ^ _ ^

Answer:

An egg with a thick shell.

Explanation:

Hope this helps.

Answer:

I think C

Explanation:

If im right give brainliest :)

The magnitude of the frictional force working on the box is 14 N.

According to Newtons's first law, an object will continue in its state of rest or state of uniform motion unless it is acted upon by an unbalanced force. This means that motion or acceleration is caused by an unbalanced force.

Since the box is moving at constant velocity, the frictional force on the box and the force that moves the box forward are exactly equal in magnitude but opposite in direction.

Learn more: https://brainly.com/question/9352088?

Answer:

Rovers, unmanned space probes, modern instruments,

Explanation:

Answer:

Gravitational energy and kinetic energy

Answer:

Las estrellas se forman debido a la unión de pequeños elementos y partículas como el hidrógeno y el helio que experimentan una fusión nuclear en el núcleo debido a la intensa presión para formar un nuevo átomo (como dos hidrógeno que se combinan para formar un átomo de helio) que es un proceso. que requiere una gran cantidad de energía para comenzar, por lo que cada reacción de fusión produce una gran cantidad de energía que puede mantener la reacción y equilibra la fuerza gravitacional que atrae las masas de las partículas hacia el centro de la estrella

A medida que la energía de la estrella se irradia, la superficie de la estrella cede la energía en forma de calor y luz de manera que debido a la alta temperatura de las partículas de los átomos de los elementos se mueven muy rápida y energéticamente para emitir fotones rápidamente. (alta frecuencia) que se ve como el alto brillo de la estrella

Explanation:

Answer:

A. Stacking the cards in order of face value

Explanation:

A deck of playing cards' entropy would be reduced by being stacked in order of face value. Option C is correct.

The degree of randomness of the system is known as entropy.

If the system is organized its entropy is less. Stacking the cards in order of face value is an organized action.

Stacking the cards in order of face value is the action would decrease the entropy of a deck of playing cards.

Hence, option C is correct.

To learn more about entropy refer to;

https://brainly.com/question/13146879

#SPJ2

Answer:

1) θ = 45,  b)  R₄₀ = R₅₀ = (v₀² / g) 0.985,  R₂₅ = R₆₅ = (v₀² / g) 0.766

3)  θ = 90º, d)   y₂₅ = (v₀ / 2g)   0.179, y₄₅ = (v₀ / 2g) 0.5, y₆₅ = (v₀ / 2g) 0.821

Explanation:

This is a projectile throwing exercise, let's use the relations

         R = [tex]\frac{v_o^2 \ sin \ 2\theta }{g}[/tex]

1) the angle for a maximum reach (R) is θ = 45

         R = v₀²/g

2) let's compare the scopes

θ = 25º

          R₂₅ = (v₀² /g) sin 2 25

          R₂₅ = (v₀² /g)  0.766

θ = 65º

         R₆₅ = (v₀² / g)  sin 2 65

         R₆₅ = (v₀² / g) 0.766

θ = 40º

        R₄₀ = (v₀²/g)  sin 2 40

        R₄₀ = (v₀²/ g) 0.985

θ = 50

        R₅₀ = (v₀²/g)  sin 2 50  

        R₅₀ = (v₀² / g) 0.985

we can see that it has the same scope

         R₄₀ = R₅₀ = (v₀² / g) 0.985

         R₂₅ = R₆₅ = (v₀² / g) 0.766

3) the maximum height is reached when the vertical speed is zero

         [tex]v_y^2 = v_{oy}^2 - 2 g y[/tex]

          v_y = 0

         y = [tex]\frac{v_{oy}^2}{2g}[/tex]

         y = (v₀² /2g) ( sin θ)²

therefore the height is maximum for θ = 90º

4) θ = 25º

         y₂₅ = (v₀ / 2g)   (sin 25)²

         y₂₅ = (v₀ / 2g)   0.179

θ = 45º

         y₄₅ = (v₀ / 2g) (sin 45)²

         y₄₅ = (v₀ / 2g) 0.5

θ = 65º

         y₆₀ = (v₀ / 2g) (sin 65)²

         y₆₅ = (v₀ / 2g) 0.821

5) It is appreciated that as the launch angle increases the height increases, the maximum is when the movement is vertical

For the launch angle, the reach has a maximum for the angle of 45º, this is due to the symmetry of the sine function around this value.

Answer:

9000 SECONDS

Explanation:3 hours=180 mins

                     180+30=210mins

                      150x60=12600secs

Answer:

9 times

Explanation:

luminosity decreases to the square of the distance

For 1000 seconds , a current of 0.1 A flows so as to transfer charge of 100C

Answer:

DONORS: If the material for which it substitutes has more electrons than the original

ACCEPTORS: If the replacement material has fewer electrons than the original material

Fermi level: the point where the probability of finding the last electron is ½

Explanation:

When in a semiconductor material a small fraction of an element is replaced by another with different valences, an excess charge is created.

If the material for which it substitutes has more electrons than the original, there is an excess of electrons, these excess electrons are weakly bound in the material and their orbits are large, in an energy versus moment diagram their energy places them a little more below the conduction band, these materials are called DONORS.

If the replacement material has fewer electrons than the original material, one electron is missing to complete the bonds, so there is a movement of the other electrons, an easier way to analyze this movement of the (n-1) electrons is to suppose that The missing charge has a positive charge and to study its movement, this positive charge is called a hole, its binding energy is small so the orbit of the hole is large, in an energy diagram it is located a little above the band of valence, these are called ACCEPTORS

The Fermi level is defined as the point where the probability of finding the last electron is ½, when the temperature is changed the density of states of the bands changes, therefore the location point moves, but its [probability remains ½