Which object has the most kinetic energy? A. Object 1 B. Object 2 C. Object 3 D. Object 4

A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef

bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0

4 years ago

Contrast the atmosphere with the biosphere

4vir4ik [10]

The biosphere contains all the planet's living things. This sphere includes all of the microorganisms, plants, and animals of Earth. Within the biosphere, living things form ecological communities based on the physical surroundings of an area. These communities are referred to as biomes. Deserts, grasslands, and tropical rainforests are three of the many types of biomes that exist within the biosphere.

The atmosphere contains all the air in Earth's system. It extends from less than 1 m below the planet's surface to more than 10,000 km above the planet's surface. The upper portion of the atmosphere protects the organisms of the biosphere from the sun's ultraviolet radiation. It also absorbs and emits heat. When air temperature in the lower portion of this sphere changes, weather occurs. As air in the lower atmosphere is heated or cooled, it moves around the planet. The result can be as simple as a breeze or as complex as a tornado.

4 0

4 years ago

A 7.45 nC charge is located 1.66 m from a 4.22 nC point charge. (a) Find the magnitude of the electrostatic force that one charg

BaLLatris [955]

Answer:

a. $F=1.03\times 10^{-7}N$

b.Repulsive

Explanation:

We are given that

$q_1=7.45nC=7.45\times 10^{-9}C$

$1nC=10^{-9}C$

$q_2=4.22nC=4.22\times 10^{-9}C$

r=1.66m

We know that

Electrostatic force = $F=k\frac{q_1q_2}{r^2}$

r=Distance between $q_1\;and\;q_2$

k=Constant= $9\times 10^9Nm^2C^{-2}$

Using the formula

a.The magnitude of the electrostatic force= $F=\frac{9\times 10^9\times 7.45\times 10^{-9}\times 4.22\times 10^{-9}}{(1.66)^2}$

The magnitude of the electrostatic force= $F=1.03\times 10^{-7}N$

b.Both charge are positive .We know that when like charges repel each other and unlike charges attract to each other.

Therefore, the force between given charges is repulsive.

7 0

3 years ago

A heavy rope, 60 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet bel

Nataliya [291]

Answer:

Riemann sum

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

Integral = W = ∫⁶⁰₀ 0.5x dx

Workdone in pulling the entire rope to the top of the building = 900 lb.ft

Riemann sum for pulling half the length of the rope to the top of the building

W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

Integral = W = ∫⁶⁰₃₀ 0.5x dx

Work done in pulling half the rope to the top of the building = 675 lb.ft

Step-by-step explanation:

Using Riemann sum which is an estimation of area under a curve

The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.

The weight of rope in that part would be 0.5Δx.

Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx

So, the Riemann sum for this total work done would be

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

The Riemann sum can easily be translated to integral form.

In integral form, with the rope being 60 ft long, we have

W = ∫⁶⁰₀ 0.5x dx

W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft

b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft

Just like in (a)

But the Riemann sum will now be from the start of the curve, to it's middle

Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

W = ∫⁶⁰₃₀ 0.5x dx

W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft

Hope this Helps!!!

7 0

3 years ago

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the alors w

Triss [41]

Answer:

A) λ = 4.88 10² nm, B) λ = 4.08 10² nm

Explanation:

The spectrum of hydrogen is correctly explained by the Bohr model, where the energy of each level is

Eₙ = -13.606 /n² [eV]

the transition generally occurs from a given level to a lower state nf <no, so a transition is

ΔE = E_f -Eₙ = -13,606 ( $\frac{1}{n_f^2} - \frac{1}{n_o^2}$ )

to find the wavelength let's use the planck relation

ΔE = h f

the speed of light is

c = λ f

we substitute

ΔE = h c /λ

λ = $\frac{h \ c}{ \Delta \lambda}$

let's apply this equation to our case

the Balmer series has as final state the level n_f = 2

A) initial state n₀ = 4, final state n_f = 2

ΔE = -13.606 ( $\frac{1}{2^2} - \frac{1}{4^2}$ )

ΔE = 2.55 eV

let's reduce to SI units

ΔE = 2.55 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.08 10⁻¹⁹ J

we calculate

λ = 6.63 10⁻³⁴ 3 10⁸ / 4.08 10⁻¹⁹

λ = 4.875 10⁻⁻⁷ m

we reduce to nm

λ = 4.875 10⁻⁷ m (10⁹ nm / 1m)

λ = 487.5 nm

we reduce to three significant figures

λ = 4.88 10² nm

B) initial state n₀ = 5

ΔE = -13,606 ( $\frac{1}{2^2} - \frac{1}{5^2}$ )

ΔE = 2,857 eV

we repeat the process of the previous point

ΔE= 2,857 1.6 10⁺¹⁹ = 4.286 10⁻¹⁹J

we look for the wavelength

λ = 6.63 10⁻³⁴ 3 10⁸ / 4.88 10⁻¹⁹

λ = 4.0758 10⁻⁷ m

we reduce to nm

λ = 4.0758 10² nm

ignificant numbers

λ = 4.08 10² nm

4 0

3 years ago