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wel
2 years ago
5

Find the recoil velocity of a 65kg ice hockey goalie who catches a 0.15kg hockey puck slapped at him at a velocity of 50m/s. Ass

ume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible.
a. −0.12m/s
b. 0m/s
c. 0.12m/s
d. 7.5m/s
Physics
1 answer:
dimulka [17.4K]2 years ago
7 0

Answer:

a. −0.12m/s

Explanation:

Given;

mass of the hockey goalie, m₁ = 65 kg

mass of the hockey puck, m₂ = 0.15 kg

velocity of the hockey puck, u₂ = 50 m/s

initial velocity of the hockey goalie, u₁ = 0

let the recoil velocity of the goalie = v

Apply Newton's third law of motion;

action and reaction are equal and opposite.

m₁v = -m₂u₂

65v = -(0.15 x 50)

65v = -7.5

v = - 7.5 / 65

v = -0.12 m/s

This shows that the goalie will move backwards with a velocity of 0.12 m/s.

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A ship leaves port and travels 50 miles at a standard position of 45⁰. The ship
laiz [17]

Answer:

   t = 4.468 h

Explanation:

For this exercise let's start by calculating the distance where the boat is

the first trip is 50 miles at 45 °, then 90 miles at 80 °,

to find the total distance let's find the distance of each displacement

             cos 45 = x₁ / 50

             sin 45 = y₁ / 50

             x₁ = 50 cos 45 = 35.35 miles

             y₁ = 50 cos 45 = 35.35 miles

             

             cos 80 = x₂ / 90

             sin 80 = y₂ / 90

             x₂ = 90 cos 80 = 15.63 miles

             y₂ = 90 sin 80 = 88.63 miles

let's find the total displacement in each axis

            x_total = x₁ + x₂

            x_total = 35.35 + 15.63

            x_total = 50.98 miles

         

            y_total = y₁ + y₂

            y_total = 35.35 + 88.63

            y_total = 123.98 miles

Let's use the Pythagorean theorem to find the modulus of the displacement

            R = √ (x_total² + y_total²)

            R = √ (50.98² + 123.98²)

            R = 134.05 miles

The boat goes at a constant speed,

           v = R / t

             

           t = R / v

let's calculate

           t = 134.05 / 30

            t = 4.468 h

8 0
3 years ago
Which of these describes the graviational force from a planet
Darina [25.2K]

An who knows

Explanation:

hmmmmm

3 0
2 years ago
A 2.0 g particle moving at 5.2 m/s makes a perfectly elastic head-on collision with a resting 1.0 g object.
sesenic [268]

Answer:

(a) The speed of the first particle is 1.75 m/s. The speed of the second particle is 6.9 m/s after the collision.

(b) The speed of the first particle is 3.45 m/s in the negative direction. The speed of the second particle is 1.73 m/s.

(c) The final kinetic energy of the incident particle in part (a) and part(b) is 0.0031 J and 0.011 J, respectively.

Explanation:

(a)

In an elastic collision, both momentum and energy is conserved.

\vec{P}_{initial} = \vec{P}_{final}\\m_1v_1 = m_1v_1' + m_2v_2'\\K_{initial} = K_{final}\\\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2

Combining these equations will give the speed of the second particle.

v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+1}(5.2) = 6.9~m/s

We can use this to find the speed of the first particle.

m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (1)(6.9)\\v_1' = 1.75~m/s

(b)

If m_2 = 10g.

v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+10}(5.2) = 1.73~m/s

m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (10)(1.73)\\v_1' = -3.45~m/s

The minus sign indicates that the first particle turns back after the collision.

(c)

The final kinetic energy of the particle in part (a) and part (b) is

K_a = \frac{1}{2}m_1v_1'^2 = \frac{1}{2}(2\times10^{-3})(1.75)^2 = 0.0031 ~J\\K_b = \frac{1}{2}m_2v_1'^2 = \frac{1}{2}(2\times10^{-3})(3.45)^2 = 0.011~J

8 0
3 years ago
A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1
Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = \rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = \frac{\rho dx}{A}

dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}

Taking the integral of both sides;we have:

\int\limits^R_0  dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx

R = 3.185 [x + \frac {x^3}{3}}]^2__0

R = 3.185 [2 + \frac {2^3}{3}}]

R = 14.863 Ω

Since V = IR

I = \frac{V}{R}

I = \frac{17}{14.863}

I = 1.144 A

∴  the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0
3 years ago
Amy wants to know whether or not an item will float when placed in a fluid. Which of the comparisons below, when true, will mean
o-na [289]

A. The object is made of stone or metal, which always sink.

B. The force of gravity is stronger than the buoyant force.

C. The buoyant force is stronger than the force of gravity.

D. The mass of the item is greater than its gravity

The answer is C

4 0
2 years ago
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