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3 years ago

A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2 m/s to 4 m/s?

Physics

2 answers:

valentinak56 [21]3 years ago

5 0

Momentum = (mass) x (speed)

Choice-B is the correct one.

mash [69]3 years ago

3 0

Answer:

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

Explanation:

The momentum of an object is given by the product between its mass (m) and its velocity (v):

$p=mv$

Let's apply this formula to calculate the initial momentum and final momentum of the ball:

- initial momentum:

$p_i = m v_i = (6 kg)(2 m/s)=12 kg m/s$

- Final momentum:

$p_f = m v_f = (6 kg)(4 m/s)=24 kg m/s$

So, the correct answer is

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

3 years ago

Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring

arlik [135]

This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

$E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J$

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

$E_k = \frac{1}{2}mv^2$

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

$E_p = 2E_k=mv^2$

From this we can determine the speed of the mass:

$E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}$

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

4 0

3 years ago

A lens of focal length 10.0 cm

fgiga [73]

Answer:

30cm

Explanation:

7 0

2 years ago

If one stand of DNA reads as AATTCCGGATCG, what would the opposite strand bases be?

3241004551 [841]

TTAAGGCCTAGC

hope this helps

4 0

3 years ago

What is the specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene v

vazorg [7]

The specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene vapor at 45 c and 1.27 atm absolute pressure will be 0 kJ/mol.

<h3>What is specific enthalpy and how was it calculated in the question?</h3>

A thermodynamic system has a property called enthalpy (H). It is calculated by the sum of the internal energy (U) of the thermodynamic system and the product of its volume (V) and pressure (p). The SI Unit is Joule (J).

Equation:

H = U+pV

The specific enthalpy of vapor can be defined as the amount of energy spent in order to transform a liquid substance into its vapor or gaseous form. The SI Unit is kJ/mol.

In the above question, the formula to be used is

P1/P2 = (Δ Hvap)/R)(1/T2-1/T1)

T1 & P1 --> the starting temperature & pressure respectively (= 1.27 atm and 45c),

T2 & P2 --> the final temperature & pressure respectively (= 0.7 atm and 45c),

R --> the real gas constant i.e. 8.314kJ/mol and

ΔHvap --> The specific enthalpy of vaporization.

Putting the values in the equation;

1.27/0.7=(ΔHvap/8.314)(1/45-1/45)

Hence as after subtracting the equation becomes 0, our final answer also comes out to be ΔHvap= 0 kJ/mol.

To know more about specific enthalpy, visit:

brainly.com/question/16244647

#SPJ4

6 0

2 years ago