7. What is the velocity of an object with a distance of 90m south and a time of
5s?

A heat engine is a device that uses to produce useful work.

Definition - a device for producing motive power from heat, such as a gasoline engine or steam engine.

So..

If this is a true or false question.. Your answer is:

TRUE

Answer:

HEAT

Explanation:

Answer:

If you mean by temperature

Water, cause say the water just sits there, it'll take a few minutes to cool down. But for concrete having all the hot sun rays on it during summer, can make it very hot, if you put 'cold water' there, it'll only last there for a moments (and that's by effect not waiting), if you wait for it to cooldown by itself, it'll take quite awhile.

If you mean solidify:

The stuff required to make concretes takes HOURS and HOURS to finish solidifying, unlike water being turn into ice normally only takes about half an hour or an hour (maybe 2) in the freezer (I don't know how long for ice, but not long at all in comparison to concrete).

Explanation:

Explanation was mostly part of the answer

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:

[tex]v = 2\pi\cdot r\cdot f[/tex] (1)

Where:

If we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:

[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]

[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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Answer:

the speed of the object has become constant.

Explanation:

At terminal velocity, air resistance equals in magnitude the weight of the falling object. Because the two are oppositely directed forces, the total force on the object is zero, and the speed of the object has become constant.

Answer:

Explanation:

Ignoring air resistance

Initial vertical velocity is 30sin35 = 17.2 m/s

Gravity reduces this velocity to zero in a time of

t = v/g =17.2 / 9.8 = 1.755 s

it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s

The horizontal velocity is 30cos35 = 24.57 m/s

the distance traveled horizontally is

d = vt = 24.57(3.51) = 86.298... = 86 m

Answer:

a. was zero while stationary and increased when the person stood.

Explanation:

momentum is mass times velocity.

initial velocity was zero

final velocity was NOT zero.

(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.

(b)  For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c) The time taken to exert the given force is 0.00625 m (s).

The given parameters;

(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]

(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]

Apply one-directional velocity equation:

[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]

Substitute the value of [tex]v_2[/tex] into equation (1);

[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]

Solve for [tex]v_2[/tex];

[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]

Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c)

The change in the momentum of the truck is calculated as;

[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]

The time taken to exert the given force is calculated as follows;

[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]

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Answer:

do it got a picture

Explanation:

Answer:

Explanation:

Relative to ground level it has

PE = mgh = 8(10)(5) = 400 J

The Universe is often studied by Scientist. The primary form of evidence that has led astronomers to conclude that the expansion of the universe is accelerating is the Observations of white dwarf supernovae.

The key evidence for that made scientist to talk about the expansion of the universe is accelerating comes from viewing and studying of white dwarf supernovae. It has been found that most distant stars all orbit at approximate speed as stars found about 30,000 light-years from their center.

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Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

I = 2/5 m R^2     inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

Answer:

An unbalanced force

Explanation:

Answer:

Force

Explanation:

Trust me

Answer:  Your answer Is A)

Explanation:    

Its direction of deflection shows it possitively charged

It brings one element into another by bombardment(transmutation)

Answer:

The one falling from the greatest height will have the greatest speed.

h = 1/2 g t^2        time for ball to fall distance h

h2 / h1 = t2^2 / t1^2       dividing equations

h2 / t2^2 = h1 / t1^2

Let v be the average speed (v2 = h2 / t2)

1 / t2 * v2 = 1 / t1 * v1

v2 / v1 = t2 / t1      the one taking the longest to fall has the greater av. speed

Check:

h4 / h1 = t4^2 / t1^2     or

t4 / t1 = (h4 / h1)^1/2

In this case t4 / t1 = (4 / 1)^1/2 = 2  or twice the average speed

t1 = (2 h / g)^1/2 = .2^1/2 = .447       using g = 10

t4 = (2 h / g)^1/2 = .8^1/2 = .894

v1 = 1 / .447 = 2.24 m/s average speed

v4 = 4 / .894 = 4.47    or twice the average speed

Answer:

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

Answer:

18.36 m/s

Explanation:

We can solve this using conservation of energy. The energy in the system will be conserved since there are no outside forces acting upon it so the potential energy and kinetic energy will be equal. Giving us this formula to start:

1/2mv^2=mgh

m=mass

g=gravity

h=height

v=velocity

We can start by figuring out the total height the rock travels which we can do by subtracting the height of the frisbee by the height the rock started at.

19m-1.8m=17.2m

Now we can plug in our variables to solve for velocity.

First we negate mass since its on both sides and cancels out leaving us with.

1/2v^2=gh

Plug in.

1/2v^2=(9.8)(17.2)

1/2v^2=168.56

v^2=337.12

v=18.36m/s

Answer:B

Explanation:

Answer: b

Explanation: b as it slows down and decelerates

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

 (comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

Answer:

Explanation:

The mass of an object given only the force acting on it and it's acceleration can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force in N

a is the acceleration in m/s²

From the question

f = 100 N

a = 2 m/s²

We have

[tex]m = \frac{100}{2} = 50 \\ [/tex]

We have the final answer as

Hope this helps you

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is  4.51 sec.

The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.

Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,

force = ma = μmg, putting the value,

a = - 1.175 m/sec²

now using equation of motion

v = u + at

t = 4.51 sec

So, time taken by the puck to come to rest is 4.51 sec.

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Answer:

Explanation:

500 W = 500 J/s

500 J/s(2 min(60 s/min) = 60000 J

Answer:

Explanation:

For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.

For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.

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[tex]a_c = 3.14\:\text{m/s}^2[/tex]

Explanation:

First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:

[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]

The centripetal acceleration [tex]a_c[/tex] is defined as

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

Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as

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

[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]

Answer:

I use mercury thermometer to measure the temperature above 80°C because the boiling point of alcohol is 78°C . So, it can't measure the temperature above 78°C.

Answer: The answer is hupa loopa

Explanation:

Answer:Let’s assume that, after the soccer ball is kicked and moves through its trajectory, it first makes contact with level ground a horizontal distance of 35 meters from where it was kicked. Let’s also assume that we can neglect air resistance. The time, t, that the soccer ball is in the air until it first contacts the ground can be found from the equation h = (1/2)gt^2 which can be rewritten as t = sqrt(2h/g) where h is the vertical distance the ball falls which is the height of the hill since the ball was kicked horizontally, and g is the acceleration of gravity which is 9.8 m/s^2. So t = sqrt(2(22)/9.8) = 2.12 seconds. In that time, the ball travelled 35 meters so its horizontal velocity was 35 meters/2.12 seconds = 16.5 meters/second.

Explanation:

Answer:

Explanation:

Conservation of angular momentum.

Disk            I = ½MR²

Point mass I = mR²  (boy)

Initial angular momentum

L₀ = Iω = ½MR²ω₀

Final angular momentum

L₁ = Iω = (½MR² + mR²)ω₁

as momentum is conserved, these are equal

(½MR² + mR²)ω₁ = ½MR²ω₀

                       ω₁ = ω₀(½MR²/ (½MR² + mR²))

                       ω₁ = ω₀(½M/ (½M + m))

                       ω₁ = 1.2(½(200)/ (½(200) + 50))

                       ω₁ = 1.2(⅔)

                      ω₁ = 0.8 cycles/second or 0.8(2π) = 1.6π rad/s