PHYSICS COLLEGE

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?

Answers

Answer 1

Answer:

Explanation:

The given data is as follows.

Mass, m = 62 kg, Initial speed, $v_(1)$ = 6.90 m/s

Length of rough patch, L = 4.50 m, coefficient of friction, $\mu_(k)$ = 0.3

Height of inclined plane, h = 2.50 m

According to energy conservation equation,

(Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

$K.E_(2) + U_(2) = K.E_(1) + U_(1) - W_(f)$

$(1)/(2)mv^(2)_(2) + U_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL$

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

$(1)/(2)mv^(2)_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL$

Cancelling the common terms in the above equation, we get

$(1)/(2)v^(2)_(2) = (1)/(2)v^(2)_(1) + gh - \mu_(k)gL$

= $(1)/(2)(6.90)^(2)_(1) + 9.8 m/s^(2) * 2.50 m - (0.3 * 9.8 * 4.50 m)$

= 36.055 - 13.23

= 22.825

$v_(2) = √(2 * 22.825)$

= 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.

Answer 2

Answer:

Answer:

Explanation:

mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s

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A celestial body moving in an ellipical orbit around a star

Answers

A celestial body moving in an elliptical orbit around a star is a planet.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

A solid cylindrical object has a mass of 2.0 kg, a diameterof0.10m, and a length of 0.18m. What is the moment of inertiaof the
body about an axis along the axis of thecylinder?

Answers

Answer:

I = 0.0025 kg.m²

Explanation:

Given that

m= 2 kg

Diameter ,d= 0.1 m

Radius , $R=(d)/(2)$

$R=(0.1)/(2)$

R=0.05 m

The moment of inertia of the cylinder about it's axis same as the disc and it is given as

$I=(mR^2)/(2)$

Now by putting the all values

$I=(2* 0.05^2)/(2)$

I = 0.0025 kg.m²

Therefore we can say that the moment of inertia of the cylinder will be 0.0025 kg.m².

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest. The two blocks stick together and move with a speed of vo /3. In terms of m, mass M is most nearly_____.

Answers

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

$m_1 = m$

$m_2 = M$

$v_(1i) =v_0$

$v_(2i) = 0$

Initial momentum will be

$P_i = m_iv_(1i)+m_2v_(2i)$

$P_i = mv_0$

After collision

$v_(1f) = v_(2f) = (v_0)/(3)$

Final momentum

$P_f = (m_1+m_2)((v_0)/(3))$

$P_f = (m+M)((v_0)/(3))$

From conservation of momentum

$P_f = P_i$

Replacing,

$(m+M)((v_0)/(3))=mv_0$

$(m+M)(1)/(3) = m$

$m+M=3m$

$M=3m-m$

$M=2m$

G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom?
b) Calculate the flow speed in the bathroom.
c) What is algebraic expression for the pressure in the bathroom?
d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.

Answers

Answer:

A) A₁ V₁ = A₂V₂

B) V₂ = 19 m /s

C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

D) P₂ = 1.88 atm

Explanation:

A) From the piaget's theory of conservation of volume, we can calculate the rate of flow of water from;

A₁ V₁ = A₂V₂

Where;

A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.

B) Using the formula given in A above, we obtain;

π x 1.2² x 4.75 = π x 0.6² x V₂

V₂ x 0.36 = 6.84

V₂ = 6.84/0.36

V₂ = 19 m /s

c ) To find pressure we shall apply Bernoulli's theorem in fluid dynamics;

P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

Where;

P₁ and P₂ are pressure at ground and second floor respectively

v₁ and v₂ are velocity at ground and second floor respectively

h₁ and h₂ are height at ground and second floor respectively ρ is density of water.

Thus, plugging in the relevant values to obtain;

4.1 x 10⁵ + (1/2 x 1000 x 4.75²) = P₂ + (1/2 x 1000 x 19²) + (5.2 x 1000 x 9.8)

(4.1 x 10⁵) + (0.11 x 10⁵) = P₂ + (1.8 X 10⁵) + (0.51 X 10

P₂ = 1.9 X 10⁵ N/m² = 1.88 atm

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An electron experiences a magnetic force of magnitude 4.60 x 10^-15 N when moving at an angle of 60.0° with respect to a magnetic field of magnitude 3.50 x 10^-3 T. Find the speed of the electron.