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A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact?

Answers

Answer 1
Answer:

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s
Answer 2
Answer:

Final answer:

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final velocity of the combined carts to the initial velocity of the first cart.

Explanation:

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.


After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answers

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

u_(1) = u_(g) = 2553.6 kJ/kg

v_(1) = v_(g) = 0.4625 m^(3)/kg

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, v_(2) = v_(g) = 0.4625 m^(3)/kg

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at v_(2) = v_(g) = 0.4625 m^(3)/kg and temperature T_(2) = 360^(o)C so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

 u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})

     u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)

                   = 2963.2 - 2.005

                   = 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

      Q - W = \Delta U + \Delta K.E + \Delta P.E ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

            \Delta K.E = \Delta P.E = 0

Now, equation will be as follows.

           Q - W = \Delta U + \Delta K.E + \Delta P.E

           Q - 0 = \Delta U + 0 + 0

           Q = \Delta U

Now, we will obtain the heat transfer per unit mass as follows.

          (Q)/(m) = \Delta u

         (Q)/(m) = u_(2) - u_(1)

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Final answer:

The heat transfer is 227.4 kJ per kg of water.

Explanation:

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

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A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)

Answers

Answer:

The frictional torque is \tau = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 = 0.341 \ kg

   The mass attached to the other end of the string is  m_2 = 0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 = m_1 * g

substituting values

      T_1 = 0.341 * 9.8

      T_1 = 3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 = m_2 * g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau = (T_2 * r ) - (T_1 * r )

substituting values  

     \tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )

     \tau = 0.2505 \ N \cdot m

Answer:here to earn points

Explanation:

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view?

Answers

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = (d)/(x)

⇒ \theta = tan^(-1) (d)/(x)

Now for the \DeltaBAC:

tan\theta = (d + h)/(x)

⇒ \theta = tan^(-1) (d + h)/(x)

Now, differentiating w.r.t x:

(d\theta )/(dx) = (d)/(dx)[tan^(-1) (d + h)/(x) -  tan^(-1) (d)/(x)]

For maximum angle, (d\theta )/(dx) = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = (-(d + h))/((d + h)^(2) + x^(2)) -(-d)/(x^(2) + d^(2))

(-(d + h))/((d + h)^(2) + x^(2)) = \frac{{d}{x^(2) + d^(2)}

After solving the above eqn, we get

x = \sqrt{(d)/(d + h)}

The observer should stand at a distance equal to x = \sqrt{(d)/(d + h)}

Final answer:

For optimum viewing of a painting in a gallery, an observer should position themselves a distance away from the painting calculated using Pythagoras theorem, forming a right-angled triangle with the painting and the floor. This distance can be expressed as c = √[(h/2 + d)² + (h/2)²], where h is the height of the painting and d is the height from the observer's eye to the bottom of the painting.

Explanation:

In the physics of optics, the viewer should position themselves to where they form a right-angled triangle with the ceiling and the painting leading to the best viewing experience. This is widely known as the 'normal viewing distance'.

Given that the painting has a height h and its lower edge is at a distance d above the observer's eye, the observer should stand a distance away from the wall, which can be calculated using Pythagoras' theorem in right triangles, which states that the square of the hypotenuse (c) is equal to the sum of the square of the other two sides (a and b), i.e., c² = a² + b²

Since the painting height and viewer height forms the right-angle in this case, we have: a = (h/2 + d), and b = h/2. Substituting a and b in Pythagoras equation, we can solve for c which is the required distance: c = √[(h/2 + d)² + (h/2)²]

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The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

Answers

Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth =  6 × 10²⁴ kg

r = Distance between Earth and Sun = 1.5* 10^(11)\ m

t = Time taken = 3 days

Acceleration is given by

a=(GM)/(r^2)\n\Rightarrow a=(6.67* 10^(-11)* 6* 10^(24))/((1.5* 10^(11))^2)\n\Rightarrow a=1.77867* 10^(-8)\ m/s^2

Velocity of the star

v=u+at\n\Rightarrow v=0+1.77867* 10^(-8)* 3* 24* 60* 60\n\Rightarrow v=0.00461031264\ m/s

The Sun's speed will be 0.00461031264 m/s

Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)

Answers

The formula for calculating the horizontal displacement of a horizontally launched projectile is x=v*t

A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.

The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.

Thus, the horizontal displacement of the projectile is given by the expression.

x=v*t

Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the ball?

Answers

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:
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