15. Which of the following describes the law of
conservation of energy?
A. Energy cannot be created or destroyed.
B. Energy can only be released through
transformation.
C. When energy is conserved, it always
changes form.
D. Energy increases when it is transferred
from one object to another.

The rocky object orbiting in space is most likely referred to as an asteroid.

This is referred to a rocky object which revolve around the sun and are considered too small to be called a planet.

Asteroids was observed with the use of a telescope and is around 250 km wide thereby making it the most appropriate choice.

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Answer:

600Kg

Explanation:

Force= mass × acceleration

mass= 1500/2.5

600kg

The mass of the rocket will be 600 kg.In the International System of Units (SI), the kilogram is the unit of mass.

Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.

In the International System of Units (SI), the kilogram is the unit of mass.

The given data in the problem is;

The net constant force is, F= 1500 N

The acceleration is,a=2. 5 m/s²

The mass of the rocket is,m=?

[tex]\rm F= mass \times acceleration \\\\ m = \frac{F}{a} \\\\ m= \frac{1500}{2.5}\\\\ m = 600 \ kg[/tex]

Hence the mass of the rocket will be 600 kg.

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A metal bar is in the xy-plane with one end of the bar at the origin, [tex]\[r = (3.87 \, \text{m}) \mathbf{i} + (3.21 \, \text{m}) \mathbf{j}\][/tex], [tex]\theta\approx 55.16^\circ\][/tex], magnitude of torque is 22.58 Nm.

a) Position Vector r can be calculated as:

Given:

[tex]\(x = 3.87 \, \text{m}\)[/tex]

y = 3.21 m

The position vector r for the point where the force is applied is:

[tex]\[r = (3.87 \, \text{m}) \mathbf{i} + (3.21 \, \text{m}) \mathbf{j}\][/tex]

b) Magnitude of Torque can be calculated as:

Magnitude of r:

[tex]\[|r| = \sqrt{(3.87 \, \text{m})^2 + (3.21 \, \text{m})^2} \approx 5.019 \, \text{m}\][/tex]

Magnitude of F:

[tex]\[|F| = \sqrt{(6.82 \, \text{N})^2 + (-3.24 \, \text{N})^2}\\\\ \approx 7.610 \, \text{N}\][/tex]

Calculate the dot product of r and F:

[tex]\[r \cdot F = (3.87 \, \text{m}) (6.82 \, \text{N}) + (3.21 \, \text{m}) (-3.24 \, \text{N})\\\\ \approx 16.658 \, \text{N} \cdot \text{m}\][/tex]

Calculate the angle [tex]\(\theta\)[/tex] using the dot product and magnitudes:

[tex]\[\cos(\theta) = \frac{r \cdot F}{|r| \cdot |F|} \\\\= \frac{16.658 \, \text{N} \cdot \text{m}}{(5.019 \, \text{m})(7.610 \, \text{N})}\\\\ \approx 0.5526\][/tex]

[tex]\[\theta \approx \cos^{-1}(0.5526)\\\\ \approx 55.16^\circ\][/tex]

Now calculate the magnitude of the torque using the sine of the angle:

[tex]\[|\tau| = |r| \cdot |F| \cdot \sin(\theta)\\\\ = (5.019 \, \text{m}) (7.610 \, \text{N}) \sin(55.16^\circ)\\\\ \approx 22.58 \, \text{N} \cdot \text{m}\][/tex]

c) Direction of Torque:

According to the right-hand rule, the torque points either into or out of the plane formed by r and F. Its direction is perpendicular to that plane. The sign of the angle determines the precise orientation.

In this case, since [tex]\(\theta \approx 55.16^\circ\)[/tex] is positive (counterclockwise rotation), the torque points out of the plane formed by r and F.

Thus, the direction of the torque is outward from the xy-plane.

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Your question seems incomplete, the probable complete question is:

A metal bar is in the xy-plane with one end of the bar at the origin. A force F=(6.82)i+ -3.24j is applied to the bar at the point x=3.87m , y=3.21m .

What is the position vector r for the point where the force is applied?

a Express your answer in terms of the unit vectors i and j .

b What are the magnitude of the torque with respect to the origin produced by F?

c What are direction of the torque with respect to the origin produced by F?

Centripetal acceleration

#1

#2

Direction of force also towards centre

Answer:

(D) light deflected around large objects because of their gravity

I hope this helps you

:)

Answer:

The correct answer is Uranus and Neptune

Gravity

when liquid splashed and is at the top, the force bringing the liquid back down to the midpoint is gravity. This brings the liquid back to its equilibrium level.

Answer:

Gravity

Explanation:

The splashing of liquid resembles like an oscillation .

The force acting on liquid to pull it down is gravity and the force bouncing it back is caused by normal reaction

Answer:

90 km/h

Explanation:

we just divide the distance by the time to get the speed.

Answer:

ω = dΘ / dt

dΘ / dt = d (3 t^2 + 2 t) / dt = 6 t + 2

At 4 sec       ω = 6 * 4 + 2 = 26 / sec

V = ω * R = 26 / sec * 2 m = 52 m/s

Answer:

b

Explanation:

the kidney eliminates waste from the blood hence no kidney will result in accumulation of waste

Answer:

15.61 m/s^2

Explanation:

From 3 rd equation of motion

s=ut+1/2at^2

given s = 30

u= 0 ( as body was at rest)

t= 1.96s

30=0*1.96+1/2a(1.96)^2

30=0+1/2a(1.96)^2

a=2*30÷(1.96)^2

a= 15.61 m/s^2 (approx)

Answer:

200 billion to 400 billion stars in a typical spiral

Explanation:

hope this helps

have an awesome day -TJ

The necessary light intensity of the disk if the disk has mass 3.89 g and radius 2.40 cm is 6.483 x 10^12cd.

a. Given that:

I = P/A

= Force x Speed/Area

Gravitational force, F= GM1M2/r^2

F= ma, F= mg

I= mg x c/q

That is,

i= mass x gravitational acceleration x speed/area

b.

Mass= 3.89g

Radius, r= 2.40cm

i=?

i= mg x c/a

Speed of light= 3.0 x 10^8 m/s

Area, a= /pi r^2

= 3.142 x (0.024)^2

= 3.142 x 0.000576

= 0.0018m^2.

i= 3.89 x 10 x 3.0 x 10^8/0.0018

i= 6.483 x 10^12cd.

c. Light has no mass, therefore, it lacks momentum and cannot exert pressure for propulsion of materials.

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Answer:

when work is done on the system or heat comes into the system

Answer:

A, flat water

Explanation:

The voltage as a function of time for the power source oscillating at the given frequency is determined as V = 169.7sin(120πt).

The peak voltage of the source is calculated as follows;

Vrms = 0.7071V₀

120 = 0.7071V₀

V₀ = 120/0.7071

V₀ = 169.7 V

ω = 2πf

where;

f is frequency = 60 Hz

ω = 2π x 60

ω = 120π rad/s

V = V₀sin(ωt)

V = 169.7sin(120πt)

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The voltage equation is a function of a time is V = 169.7sin(120πt) if the 1. 5-kω resistor and 27. 7-mh inductors are connected in series to a vrms = 120 v ac power source oscillating at a frequency of f = 60 Hz.

Electromagnetic induction causes the induced voltage. Electromagnetic induction is the process of creating emf (induced voltage) by exposing a conductor to a magnetic field.

We have:

[tex]\rm V{rms} = 120[/tex]

We know:

[tex]\rm V_{rms} = 0.7071V_o[/tex]

[tex]\rm 120 = 0.7071V_o[/tex]

[tex]\rm V_o = 169.70[/tex] Volts

Expression for the angular frequency is given by:

ω = 2πf

Where f is frequency

ω = 2π x 60  

ω = 120π rad/s

Voltage equation as a function of time is given by:

V = V₀sin(ωt)

V = 169.7sin(120πt)

Thus, the voltage equation is a function of a time is V = 169.7sin(120πt) if the 1. 5-kω resistor and 27. 7-mh inductors are connected in series to a vrms = 120 v ac power source oscillating at a frequency of f = 60 Hz.

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For a person dragging a packing crate of mass 100 kg, the angle above the horizontal  is mathematically given as

theta = 21.8 °

Generally, the equation for the Force   is mathematically given as

[tex]Fnet = F_{costheta} - F_{friction}[/tex]

Therefore

Fnet = Fcostheta - mu*((mg) - Fsintheta)

sin theta = mu*costheta

mu = tantheta

theta = tan^-1(mu)

theta = tan^-1(.400)

theta = 21.8 °

In conclusion, the angle above horizontal

theta = 21.8 °

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Answer:

the current of 12 ohm is 2A

Answer:

55 j

Explanation:

Mechanical energy = kinetic + potential

                  65          = 10 + P.E.

                    55 j = PE

Answer:

the current of 12 ohm is 2Amps

Answer:B

Explanation:  

A has about 248 hp  but B has about 683 hp. 2016 is younger too. good day

Answer:

So you maybe don't fall, have something fall on you, and for a bunch of different reasons. EARFQUAKES are really dangrous.

Explanation:

Hope this helps you!

The magnitude of the centripetal acceleration of the bicycle wheel is 40 m/s².

The centripetal acceleration of the bicycle wheel is the inward or radial acceleration of the bicycle on the circular path.

The magnitude of the centripetal acceleration of the bicycle wheel is calculated as follows;

a = ω²r

where;

ω is angular speed

r is the radius of the circular path

a = (10)² x 0.4

a = 40 m/s²

Thus, the magnitude of the centripetal acceleration of the bicycle wheel is 40 m/s².

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The acceleration of a body travelling in a circular route is known as centripetal acceleration.  The centripetal acceleration of the wheel at this time point is 40 m/sec².

The acceleration of a body travelling in a circular route is known as centripetal acceleration. It is given by the formula,

[tex]a_c = \dfrac{v^2}{r} = \omega^2 r[/tex]

The centripetal acceleration of the wheel with angular velocity 10 rad/sec and velocity 40cm (0.4 m) can be written as,

[tex]a_c = \omega ^2\times r\\\\a_c = (10)^2\times 0.4\\\\a_c = 40\rm\ m/sec^2[/tex]

Hence, the centripetal acceleration of the wheel at this time point is 40 m/sec².

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Using the formula for  permutation, there are 5040 ways that  7 objects are taken 1 at a time.

The permutation of objects refers to the number of ways that an object can be selected from a number of objects.

Given that;

nPr = n!/(n - r)!

Hence;

7P1 = 7!/(7 - 1)!

7P1 = 7 * 6 !/6!

7P1 = 7 * 6 * 5 * 4 * 3 * 2 *1

7P1 = 5040

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An expression for the coefficient of kinetic friction between the block and the horizontal surface is given as μk = Ff/mg.

The coefficient of kinetic friction is the force of frcition between a moving object and the horizontal surface.

The expression for the coefficient of kinetic friction between the block and the horizontal surface is given as;

Ff = μkFₙ

Ff = μkmg

μk = Ff/mg

where;

Thus, an expression for the coefficient of kinetic friction between the block and the horizontal surface is given as μk = Ff/mg.

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The maximum mass of coal that the barge will carry in freshwater without sinking is 0.111 kg.

The volume of the barge is calculated as follows;

V = L³

V = (4.8 cm)³

V = 110.6 cm³

For the barge to float in freshwater, its density must be less or equal to density of water.

density(barge) = density(water)

density(barge) = 1 g/cm³

mass = density x volume

mass = 1 g/cm³ x 110.6 cm³

mass = 110.6 g

mass = 0.111 kg

Thus, the maximum mass of coal that the barge will carry in freshwater without sinking is 0.111 kg.

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(a) The impulse on the puck is 1.323 kgm/s.

(b) The magnitude of the force that acted on the puck is 264.5 N.

The impulse on the puck is equal to the change in momentum of the puck.

The change in th momentum of the puck is calculated as follows;

ΔP = m(u - v)

where;

Magnitude of the final velocity

|v| = √[(-25)² + (5)²] = 25.5 m/s

ΔP = 0.115(37 - 25.5)

ΔP = 1.323 kgm/s

Thus, the impulse on the puck is 1.323 kgm/s.

The magnitude of the force that acted on the puck is calculated as follows;

F = ΔP/t

F = 1.323/0.005

F =264.5 N

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