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1 year ago

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A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg an

d is at rest. Assuming a closed system, find the total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper. In 3-4 sentences, identify the object with the greater momentum after the puck is caught and explain your reasoning.

1 answer:

Ainat [17]1 year ago

4 0

Answer:

Explanation:

Momentum is the product of mass of a body and its velocity.

Given the mass of the puck m1 = 0.16kg

velocity of the puck v1 = 40m/s

Given the mass of the goalkeeper m2 = 120kg

velocity of the goalkeeper v2= 0m/s (goal keeper at rest)

The total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper is expressed as:

m1v1 + m2v2 (their momentum will be added since they collide)

= 0.16(40) + 120(0)

= 0.16(40) + 0

= 6.4kgm/s

Let us calculate their common velocity using the conservation of momentum formula;

m1u1 + m2u2 = (m1+m2)v

6.4 = (0.16+120)v

6.4 = 120.16v

v = 6.4/120.16

v = 0.053m/s

Hence after collision, both objects move at a velocity of 0.053m/s

Momentum of the puck after collision = m1v

Momentum of the puck after collision = 0.16*0.053m/s

Momentum of the puck after collision = 0.0085kgm/s

Momentum of the keeper after collision = m2v

Momentum of the keeper after collision = 120*0.053m/s

Momentum of the keeper after collision = 6.36kgm/s

From the calculation above, it can be seen that the keeper has the greater momentum after the puck was caught since the momentum of the keeper after collision is greater than that of the puck

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A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.

Airida [17]

Answer:

Part A : E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B : V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

F = $\frac{1}{4\pi}$ ε₀ Q₁/R₁²

Then the electric field at that point is

E = F/1

E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

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1 year ago

A certain amusement park ride consists of a large rotating cylinder of radius R=3.05 m.R=3.05 m. As the cylinder spins, riders i

aniked [119]

Answer:

a. N = 2.49W b. 0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ = mg/mrω²

= W/N

= 9.8m/24.43m

= 0.40

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1 year ago

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

2 years ago

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb

Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g

Buoyant force on lead = .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115 / 33.5

= .36

= 36 %

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1 year ago

An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.

viktelen [127]

Answer:

The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Explanation:

Given that,

Current = 8.60 A

Velocity of electron $v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s$

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu I}{2\pi d}(-k)$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}$

$B=0.0000086\ T$

$B=-8.6\times10^{-6}k\ T$

We need to calculate the force that the wire exerts on the electron

Using formula of force

$F=q(\vec{v}\times\vec{B}$

$F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )$

$F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k$

$F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Hence, The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

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2 years ago