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3 years ago

What is her velocity? 1.5 m/s 2 m/s 2.5 m/s 5 m/s

Physics

2 answers:

fiasKO [112]3 years ago

7 0

Answer:

There is no actual question attached to this, to get a real answer be sure to include the documents/question that is provided on your work.

Explanation:

Xelga [282]3 years ago

6 0

Answer:

C. 2.5

Explanation:

This is the answer to that one question on Edge nuity, I got it right.

You might be interested in

How do the wavelength and frequency of red light compare to the wavelength and frequency of blue light?

11111nata11111 [884]

Answer:

Red light has a longer wavelength and lower frequency than blue light.

Explanation:

The photon energy carried by the red light is lower than blue light. The higher the frequency of the photon, the greater its energy and the greater its wavelength, the lower its energy. Thus, red light has a longer wavelength and lower frequency than blue light.

6 0

3 years ago

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W

ArbitrLikvidat [17]

Complete question:

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.

Answer:

The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

Explanation:

Given;

power of the flux produced, P = 1 W = 1 J/s

energy per photon, E = 78 keV

Convert the energy per photon to J

E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon

let the number of photons = n

n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s

$n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s$

Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

8 0

2 years ago

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per

vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work and amount of entropy and magnitude

solution

first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1

so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )

put these value we get

entropy change = 7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum theoretical work = specific internal energy - 2038.3

maximum theoretical work = 2728 - 2038.3

maximum theoretical work is 689.4 kJ/kg

3 0

3 years ago

What do we call the quantity of matter that an object contains?

german

Answer: A MASS

Explanation:LOOKED IT UP

3 0

3 years ago

Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,

ikadub [295]

Answer:

$p_t_o_t_a_l=250kg\frac{m}{s}$

Explanation:

The total momentum of a system is defined by:

$(mv)_t_o_t=m_1v_1+m_2v_2+...$

Where,

$(mv)_t_o_t$ is the total momentum or it could be expressed also as $p_t_o_t_a_l$ .

$m_1$ and $m_2$ represents the masses of the objects interacting in the system.

$v_1$ and $v_2$ are the velocities of the objects of the system.

Remember: The momentum is a fundamental physical magnitude of vector type.

We have:

$m_1=725 kg$

$v_1=15\frac{m}{s}\\m_2=625 kg$

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be negative. It goes in a different direction from car A.

$v_2=-17\frac{m}{s}$

Then the total momentum of the system is:

$p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}$

8 0

3 years ago