You must exert a force of 5N on a book to slide it across a table.If you do 2.5 J of work in the process, how far has the book moved?

Answer:

Those are all right

Explanation:

Answer:

They are right! Nice job!

Explanation:

Answer:

I think electricity is one kind of energy

The electricity generated by centralized generation is distributed through the electric power grid to multiple end-users. Centralized generation facilities include fossil-fuel-fired power plants, nuclear power plants, hydroelectric dams, wind farms, and more.

Answer:

hola bb como estas dime bb en que te ayudo

Explanation:

Answer:

Explanation:

a ) 12.0 lbs  -  mass

b ) .34 g   - mass

c ) 120 kg    - mass  

As lbs , g and kg are units of mass , these things represent mass . lbs is FPS unit of mass and kg is  SI unit . g is CGS unit of force .

d ) 1600 kN     It represents force because kN or kilonewton is unit of  force .

e ) it represents , length or distance because m or metre  is unit of length .

f ) 411 cm It represents length or distance because cm is smaller unit of length

g ) 899 MN or million newton .  Like d ) It also represents force . Million newton is higher unit of force .

Answer:

Yes, at the highest point in the trajectory velocity and acceleration are perpendicular.

Explanation:

Let [tex]x-[/tex]axis be the ground and [tex]y-[/tex]axis is the height above the ground. Let the initial velocity of the object is [tex]u[/tex] having angle [tex]\theta[/tex] with respect to ground and the acceleration due to gravity , [tex]g[/tex], is acting in vertically downward direction as shown in figure.

As gravitational force is acting in vertical direction, so, it will not change the horizontal velocity of the object. So, at any instant throughout the projectile motion the [tex]x-[/tex]component of the initial velocity will remailns constant, i.e. [tex]u_x=u \cos \theta[/tex].

While, due gravitational force, [tex]y-[/tex]component of the initial velocity will change. Initially, [tex]u_y=u \sin \theta[/tex] is in vertically upward direction and gravitational force is actiongn vertically downward direction, so, at first [tex]u_y[/tex] will decrease untill it reaches the highest point of the trajectory as shown. At the highest point the vertical component of the velocity [tex]u_y=0[/tex], so, there is only horizontal component of the velocity. i.e [tex]u_x= u \cos\theta[/tex] .

Now, the resultant velocity of the object at the highest point is [tex]u \cos \theta[/tex] which is in the horizontal direction while the acceleration , [tex]g[/tex] , (due to gravily) is actiog in vertically downward direction.

Hence, at the highest point in the trajectory velocity and acceleration are perpendicular.

Thus the response above is appropriate.

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Answer:

a = 1.16 m/s²

Explanation:

In order to find the rate of acceleration of the ball, we will use third equation of motion, as follows:

2as = Vf² - Vi²

where,

a = rate of acceleration = ?

s = distance covered by the ball = 21.9 m

Vf = Final Velocity of the ball = 7.14 m/s

Vi = Initial Velocity of the ball = 0 m/s (Since, the ball started from rest)

Therefore,

2(a)(21.9 m) = (7.14 m/s)² - (0 m/s)²

a = (50.97 m²/s²)/(43.8 m)

a = 1.16 m/s²

Answer:

a) compresses and warms.

Explanation:

When air sinks, it falls down and compresses the air close to the surface of the ground, which raises the temperature of the air (since compression results in a rise in temperature). Also, as the air sinks, it prevent the already warm air close to the surface from rising, further increasing its own temperature, warming it up.

Answer:

Efficiency Or C. on e2020

The ratio of output work to input work of a machine expressed as a percent refers to efficiency of the machine.

How well a machine converts its input energy into useable output energy, or work, is referred to as its efficiency. It is the portion or percentage of the output divided by the input, and it plays a significant role in a machine's utility.

It denotes the degree of performance that requires the fewest inputs to provide the greatest amount of output in more mathematical or scientific words. It frequently particularly refers to the ability of a particular application of effort to deliver a particular output with the least amount of waste, expense, or superfluous work. Efficiency covers a wide range of inputs and outputs across numerous disciplines and sectors.

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Answer:

C- A mixture can vary in composition, but a pure substance has a set composition

Explanation:

The main difference between a mixture and a pure substance is: A mixture can vary in composition, but a pure substance has a set composition ( option c )

A mixture is a substance which contains two or more constituents which can easily be separated by a physical means. It consists of two or more different substances, not chemically joined together.

A pure substance consists only of one element or one compound.

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Answer:

V = (5.8cm/s)i, (4.7cm/s)j

Explanation:

Given :

r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^

To obtain the average velocity (V)

V = (r2 - r1) / (t2 - t1)

To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above

r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j

r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j

r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j

r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j

r2 = (16.1cm)i + (9.4cm)j

V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0

V = 11.6i / 2 ; 9.4j / 2

V = (5.8cm/s)i, (4.7cm/s)j

Answer:

[tex] \lim_{(x,y) \to \(0,0)} \frac{2x {y}^{2} }{ {x}^{2} {y}^{2} } [/tex]

Explanation:

[tex]\lim_{(x,y) \to \(0,0)}[/tex]

Answer:

93.615 million or more

Explanation:

The Sun is about 93 million miles from Earth to calculate the distance to a star, astronomers observe it from different places along Earth's orbit around the Sun.

2.5 better than the first one more game to go hard but I 46 the bucks

Answer:

hsiahdjjsdhishshdakyeacs4y9yssig

Explanation:

ntesgs7thsjv6wlubx4euc

Answer:

I think it would be interviewing the researchers I'm unsure though

Answer:

Probably C) Conducting the experiment again

Explanation:

Option A is wrong because there is a chance that the group of researchers made a mistake or got the wrong data. So when interviewing the researchers, there is a chance of them giving the wrong data, especially since the experiment has never been conducted before.

Option B is wrong because the experiment the researchers where doing was new and has never been conducted before. Therefore there is no older research done on that topic.

Option C is correct because when conducted the experiment again, they can compare their results with the group of researchers. Checking if the data is correct.

Option D is wrong because the group of researchers might have made a mistake when writing down the procedure, leading to false data.

Answer:

800 miles per hour

Explanation:

Answer:

13 hours, 16 minutes, 41 seconds

Explanation:

Distance of Saturn from the Sun

= 890,700,000 miles

[tex] = 8.907\times 10^8 \: miles\\

= 8.907\times 10^8 \times 1.61 \times 10^3 \:meters\\

= 14.34027\times 10^{11}\: m\\

= 1. 434027\times 10^{12}\: m\\\\

\because Time = \frac{Distance}{Speed} \\

\therefore Time = \frac{1.434027\times 10^{12}}{3\times 10^8}\\

\therefore Time = 4.78009 \times 10^{12-8}\\

\therefore Time = 4.78009 \times 10^{4} \:seconds \\

\therefore Time = 47800.9 \:seconds\\

\therefore Time = 47801 \:seconds\\

\therefore Time = 13h\: 16m\: 41s\\[/tex]

Saturn is 890,700,000 miles from the Sun. This distance in the scientific notation would be 1.434×10¹² metes.

If the speed of light is 3.0 x 108 meters/second, then it would take 4780 seconds for light from the Sun to reach Saturn.

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second. It can also be represented by the infinitesimal rate of change of displacement with respect to time. The generally considered unit for velocity is a meter per second.

As given in the problem Saturn is 890,700,000 miles from the Sun

1 mile = 1.61 x 10³ meters

890,700,000 miles = 890,700,000×1.61×10³ meters

                                =1.434×10¹² metes

As given  If the speed of light is 3.0 x 10⁸ meters/second

Time = distance /speed

        = 1.434×10¹² metes / 3.0 x 10⁸ meters/second

        = 4780 seconds

Thus, the distance in the scientific notation would be 1.434×10¹² metes and it would take 4780 seconds for light from the Sun to reach Saturn.

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Answer:

Three forms of energy when a match is being lit are potential, kinetic and thermal.

hope this helps! :)

Explanation:

Answer:

Potential Energy

Kinetic Energy

Thermal Energy

Explanation:

Answer:

stable is the answer

Explanation:

stable electrons

Answer:

26.64 m

Explanation:

Given the following :

Acceleration at ocean surface = 0.0800 m/s²

Distance covered if initial speed = 0.700 m/s and accelerates to a speed of 2.18m/s

Using the equation :

v² = u² + 2as

Where ;

v = final velocity ; u = initial velocity ; a = acceleration ; s = distance covered

Therefore,

v² = u² + 2as

2.18² = 0.7² + (2 × 0.08 × s)

4.7524 = 0.49 + 0.16s

4.7524 - 0.49 = 0.16s

4.2624 = 0.16s

s = 4.2624 / 0.16

s = 26.64 m

Answer:

Explanation:

According to Newton's first law of motion "An object will continue to stay or remain at rest or continue to be in constant motion "unless acted upon by an external force"

F= ma

Basically a force is a pull or a push

so in other words the two ways by which an object can get in motion is by

1. A pull force

2. A push force

Answer:

The man has, at most, 0.4 secs to get out of the way

Explanation:

First, we will determine, the time the block takes to reach height 13.4 m above the ground.

Also, we will determine the time it will take to reach the man, if he does not notice it.

Then, the difference in the two times will be the time the man has to get out of the way.

From the formula

[tex]h =ut - \frac{1}{2}gt^{2}[/tex]

Where [tex]h[/tex] is height

[tex]u[/tex] is the initial velocity

[tex]t[/tex] is time

and [tex]g[/tex] is acceleration due to gravity ( Take [tex]g =[/tex] 9.8 m/s²)

Let the time the block takes to reach height 13.4 m be [tex]t_{1}[/tex]

The cement block falls from a 51.1 m high building, then by the time it reaches a height 13.4 m above the ground, the displacement is

13.4 m - 51.1 m = - 37.7 m  

The is the height the block has fallen from by the time it gets to a height 13.4 m above the ground

Then,  

[tex]h =ut - \frac{1}{2}gt^{2}[/tex]

[tex]-37.7 = (0)(t_{1}) - \frac{1}{2}(9.8)t_{1}^{2}[/tex]

(NOTE: [tex]u[/tex] = 0 m/s because the cement block falls from rest)

[tex]37.7 = 4.9t_{1}^{2}[/tex]

[tex]t_{1}^{2}= \frac{37.7}{4.9}[/tex]

[tex]t_{1}= \sqrt{7.69} \\[/tex]

[tex]t_{1}= 2.77 secs[/tex]

Hence, It takes the block 2.77 secs to reach height 13.4 m

Also,

Let the time it will take to reach the man be [tex]t_{2}[/tex]

The man is 1.70 m tall, hence, to reach the man, the cement block would have reached a height ( 1.70 m - 51.1 m) = - 49.4 m

Then,

[tex]h =ut - \frac{1}{2}gt^{2}[/tex]

[tex]-49.4 = (0)(t_{2}) - \frac{1}{2}(9.8)t_{2}^{2}[/tex]

[tex]49.4 = 4.9t_{2}^{2}[/tex]

[tex]t_{2}^{2} = \frac{49.4}{4.9}[/tex]

[tex]t_{2}= \sqrt{10.08}[/tex]

[tex]t_{2}= 3.17 secs[/tex]

Hence, it will take 3.17 secs to reach the man

Now, for the time the man has to get out of the way, that is

[tex]t_{2} - t_{1}[/tex] = 3.17 secs - 2.77 secs

= 0.4 secs

Hence, the man has, at most, 0.4 secs to get out of the way

Explanation:

Given:

v₀ = 15.68 m/s

a = -9.8 m/s²

t = 0.4n s, n = 0, 1, 2, etc.

Find: Δy and v

To find displacement, use the equation:

Δy = v₀ t + ½ at²

To find velocity, use the equation:

v = at + v₀

[tex]\left\begin{array}{ccc}t&y&v\\0&0&15.68\\0.4&5.49&11.76\\0.8&9.41&7.84\\1.2&11.76&3.92\\1.6&12.54&0\\2.0&11.76&-3.92\\2.4&9.41&-7.84\\2.8&5.49&-11.76\\3.2&0&-15.68\end{array}\right[/tex]

The ball lands after 3.2 seconds.

Answer:

calculated velocities:

(t,v(t))

(0.4,11.756)

(0.8,7.832)

(1.2,3.908)

(1.6,-0.016)

calculated displacements:

[tex](t,\Delta x)[/tex]

(0.4, 5.487)

(0.8,9.4048)

(1.2, 11.7528)

(1.6, 12.5312)

the ball spends 3.2 seconds in the air

Explanation:

Assuming the ball is launched on earth with standard gravity and that the effects of drag is neglected.

assume that 'up' is positive direciton since that is direction of travel

'down' is negative direction

formula for velocity is [tex]v(t)=v_0+at[/tex]

where [tex]v(t)[/tex] is the velocity at time [tex]t[/tex]

[tex]v_0[/tex] is the initial velocity at time [tex]t=0[/tex]

[tex]a[/tex] is the acceleration in the direction of the velocity

velocity is in m/s and accelration is in m/s^2 and t is in seconds

for displacement: [tex]\Delta x=v_0 t+\frac{1}{2}at^2[/tex]

Known stuffs:

in our case, we know we are on earth so, [tex]a=-9.81[/tex]m/s^2. the negative sign indicates that the direction is oposite the velocity since the ball travels up but gravity pulls down

the initial velocity is given as [tex]v_0=15.68[/tex]m/s

therefore, our equation for velocity is

[tex]v(x)=15.68-9.81t[/tex]

now we just need to plot it every 0.4 seconds

this is easy subsituteion. here's the first:

[tex]v(0.4)=15.68-9.81(0.4)=11.756[/tex]m/s

if we keep plotting this, we get the following table

(t,v(t))

(0.4,11.756)

(0.8,7.832)

(1.2,3.908)

(1.6,-0.016) the velocity is negative meaning that the ball is now going 'down' and therefor the ball has just passed the peak of it's upward travel

so since it took about 1.6 seconds to get here, it will spend about twice this time in the air or about 3.2 seconds

to get an exact answer, solve for the time when velocity is 0, ie when the ball has reached its peak and stopped moving

set [tex]v(t)=0[/tex] and solve for t

[tex]0=15.68-9.81t\rightarrow 15.68=9.81t\rightarrow t=1.598s[/tex]

so it takes 1.598 seconds to get to peak, therefore it spends twice this time in the air or 3.197 seconds in the air. it's close enough to 3.2 seconds

the ball spends 3.2 seconds in the air

displacement:

[tex]\Delta x=v_0t+\frac{1}{2}at^2[/tex]

[tex]\Delta x=15.68t+\frac{1}{2}(-9.81)t^2[/tex]

some values are

[tex](t,\Delta x)[/tex]

(0.4, 5.487)

(0.8,9.4048)

(1.2, 11.7528)

(1.6, 12.5312)

Answer:

Explanation:

Given the capacitance of a capacitor = 8.5F

Voltage across the capacitor = 2te^-3tV

The current across a capacitor is expresses as shown;

I = Cdv/dt

Given V = 2te^-3tV

dv/dt = 2t(-3e^-3t) + 2(e^-3t)

dv/dt = -6te^-3t + 2e^-3t

Factor out the common term

dv/dt = -2e^-3t{3t-1}

If C = 8.5F

I = 8.5 × -2e^-3t{3t-1}

I = -17.0e^-3t{3t-1} A

Power is the product of current and voltage.

P = IV

Given I = -17.0e^-3t{3t-1} A

V = 2te^-3t V

Power = -17.0e^-3t{3t-1} × 2te^-3t

Power = (17e^-3t - 51te^-3t) × 2te^-3t

Power = 34te^{-3t-3t} - 102t²e^{-3t-3t}

Power = 34te^-6t - 102t²e^-6t

Power = 34te^-6t{1-3t} Watts

Answer:

An applied force is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object.

Explanation: