All categories

3 years ago

Calculate the momentum of a 0.145 kg baseball being thrown at a speed of 40.0 m/s.

Physics

1 answer:

Savatey [412]3 years ago

3 0

Answer:

5.8 kg*m/s

Explanation:

p = mv

momentum = mass times velocity

momentum = 0.145 kg times 40.0 m/s

momentum = 0.145* 40.0

momentum = 5.8 kg*m/s

You might be interested in

A box of mass 10 kg is pulled from the hold of a ship with an acceleration of 1 m/s^2 by a vertical rope attached to it .find th

wlad13 [49]

F=ma
Mass times acceleration
We have g (10ms^_2) and a (1 given)
So total would be
10 kg times (10+1) =
110 N

5 0

3 years ago

A sound wave has a frequency of 265 hz and a wavelength measured at 1.3

pochemuha

Answer:

V = f λ speed of wave in terms of frequency and wavelength

t = S / V time for wave to travel a distance S

t = 91.4 m / 344.5 m/s = .265 sec time to travel 91.4 m

8 0

2 years ago

Find the frequency of a wave with the wavelength 3.5 m and the speed is 50 m/s. <br>

Andru [333]

Answer:

8.57 Hz

Explanation:

From the question given above, the following data were obtained:

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

The velocity, wavelength and frequency of a wave are related according to the equation:

Velocity = wavelength × frequency

v = λ × f

With the above formula, we can simply obtain the frequency of the wave as follow:

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

v = λ × f

30 = 3.5 × f

Divide both side by 3.5

f = 30 / 3.5

f = 8.57 Hz

Thus, the frequency of the wave is 8.57 Hz

7 0

3 years ago

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago

For a cylindrical capacitor, the capacitance does not depend on which of the following values?

IgorC [24]

Answer:

Capacitance of cylindrical capacitor does not depends on the amount of charge on the conductors

Explanation:

Consider a cylindrical capacitor of length L, inner radius R₁ and outer radius R₂, permitivity ε₀ constant then capacitance of cylindrical capacitor is given by:

$C=\frac{2\pi \epsilon_{o}L}{ln\frac{R_{2} }{R_{1}} }$

From this equation it is clear that capacitance of cylindrical capacitor is independent of the amount of charge on the conductors where as directly proportional permitivity constant and length of cylinder where as inversely proportional to natural log of ratio of R₂ and R₁

5 0

3 years ago