If your water heat has an efficiency of 95 percent, how much energy would it take to heat 45kg of water from 23 C to 60 C. (The specific heat of water is 4.18 J/g/C.) Please show your work

Answer:

2.2 ma

Explanation:

Given :

Length of the rope = L

Mass of the rope = m

Mass of the object pulled by the rope = M

M = 1.5 m

So, L [tex]$\rightarrow$[/tex] m

For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]

∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.3 m

And for remaining 0.7 L =  [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.7 m

By Newtons law of motion,

F - T = ( 0.3 m) a .........(1)

T = ( M + 0.7 m) a

T = ( 1.5 m + 0.7 m) a

T = ( 2.2 m ) a  ..............(2)

So from equation (1) and (2), we have

Tension on the rope

T = 2.2 ma

Answer: Option b.

Explanation:

We know:

Wavelength = 632.8 nm

Fluence = 4.5*10^20 photon/s

The energy of a single photon of wavelength λ is:

E = (h*c)/λ

where:

h = 6.6*10^(-34) J*s

c = 3*10^8 m/s

And we should rewrite the wave length in meters, so:

λ = 632.8 nm = 632.8*10^(-9) m

replacing these in the energy equation, we get:

E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J

So each one of the  4.5x10^20 photon that flow each second have this energy, then the power is:

P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s

and 1 W = 140.85 J/S

Then the power is:

P = 140.85 W

Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.

Answer:

Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.

Explanation:

The speed of the hoop when it has rolled halfway up the side of the pipe is √(v₀² - gR).

The speed of the hoop when it has rolled halfway up the side of the pipe is calculated as follows;

K.E = P.E

- ¹/₂mv₀²  + ¹/₂Iω² = (mgh₀ - mghf)

- ¹/₂mv₀²  + ¹/₂Iω² = (0 - 0.5mgh)        (hf = 0.5h) (half way up)

¹/₂Iω² = ¹/₂mv₀² - 0.5mgh  

where;

¹/₂(mr²)(vf/r)² = ¹/₂mv₀²  - 0.5mgh

¹/₂vf²  = ¹/₂v₀² - ¹/₂gh  

vf²  = v₀² - gh

vf = √(v₀² - gh)

where;

vf = √(v₀² - gR)

[tex]v_f = \sqrt{v_0^2 - gR}[/tex]

Learn more about moment of inertia of hoop here: https://brainly.com/question/14956994

Answer:

9m/s^2

Explanation:

F=ma

a=F/m

a=63/7

a=9m/s^2

Answer:

Look Below -->

Explanation:

a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )

b. Her displacement is  0 km because she went back home. (Displacement is the difference between the end and starting points)

c. 3 km/hr (30 minutes / 10 km)

Explanation:

Projectile motion, Bernoulli's principle, and Magnus Effect.

Sure I would be happy to discuss projectile motion!

I'll do it if you mark brainliest :) I need the points thanks

Answer:

Potential Energy

Explanation:

Answer:

momentum = mass * velocity ... kg•m/s

acceleration = v / t = 200 m/s / 11 ms ... m/s^2

force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N

Explanation:

Answer:

c

Explanation:

Answer:

towards the north.

Explanation:

Answer:

μs = 0.36

Explanation:

       [tex]F_{frmax} = \mu_{s}* F_{n} = m * g (1)[/tex]

      [tex]F_{n} = m* \omega^{2} * r (2)[/tex]

       [tex]\mu_{s} = \frac{g}{\omega^{2} r} (3)[/tex]

       [tex]\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)[/tex]

Answer:

a) the weight of the person is 170.5 lbf

b) weight of the astronaut on the moon is 28.2 lbf

Explanation:

Given the data in the question;

a)

we know that;

weight on the surface of the earth = m[tex]g_{earth[/tex]

given that m = 170.5 lbm and g = 32.174 ft/s²

we substitute

weight on the surface of the earth =  170.5 lbm × 32.174 ft/s²

= 5485.667 lbm-ft/s²

1 lbf = 32.174 lbm-ft/s²

so

weight on the surface of the earth = (5485.667 / 32.174) lbf

weight on the surface of the earth = 170.5 lbf

Therefore, the weight of the person is 170.5 lbf

b)

given that;

weight on the surface of the earth = m[tex]g_{moon[/tex]

m = 170.5 lbm and g = 5.32 ft/s²

weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²

= 907.06 lbm-ft/s²

1 lbf = 32.174 lbm.ft/s²

weight on the surface of the earth = ( 907.06 / 32.174 ) lbf

weight on the surface of the earth = 28.2 lbf

Therefore, weight of the astronaut on the moon is 28.2 lbf

Answer:

20.49 N

Explanation:

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

Power (P) = 0.824 hp

Velocity (v) = 30 m/s

1 h = 746 W

Force (F) =?

Next, we shall convert 0.824 hp to Watts (W). This can be obtained as follow:

1 h = 746 W

Therefore,

0.824 hp = 0.824 hp × 746 W / 1 h

0.824 hp = 614.704 W

Finally, we shall determine the force as follow:

Power (P) = 614.704 W

Velocity (v) = 30 m/s

Force (F) =?

P = F × v

614.704 = F × 30

Divide both side by 30

F = 614.704 / 30

F = 20.49 N

Answer:

[tex]a=62.5\ m/s^2[/tex]

Explanation:

Given that,

The velocity of the object, v = 25 m/s

Radius of the circle, r = 10 m

We need to find the acceleration of the object. The centripetal acceleration act on it. It is given by :

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(25)^2}{10}\\\\a=62.5\ m/s^2[/tex]

So, the acceleration of the object is equal to [tex]62.5\ m/s^2[/tex].

Answer:

An object is considered to be in motion only when its position changes over time with reference to a point which will be known as orgin

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Waves.

Wavelength is the distance between two consecutive crest or trough.

hence, here the distance is 10cm

So the wavelength is 10cm

===> 10 cm

Answer:

1.67 m/s²

Explanation:

Total force is;

ΣF = F_f + F_a

From the free body diagram attached, we see that the frictional force F_f is acting in a negative direction.

Thus;

ΣF = -15 + 30

ΣF = 15 N

Now, to get the acceleration, we will use the formula;

ΣF = ma

We are given m = 9 kg

Thus;

15 = 9a

a = 15/9

a = 1.67 m/s²

Answer:

20 seconds

Explanation:

Answer:

A. Protons

Explanation:

Atomic number is an indicator of how many protons an element possesses in its nucleus. We use atomic number to identify an element. For example, carbon has 6 protons and an atomic number of 6.

Answer:

Explanation:

From the given information:

At state 1:

Initial Quality [tex]= x_1 = 0.85[/tex]

mass = 10.0 kg

At state 2:

Temperature [tex]T_2 = 320^0[/tex]

mass of the piston [tex]m_p = 204 \ kg[/tex]

area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]

Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]

Gravitational acceleration = 9.81 m/s²

[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.

To calculate the applying force balance over the piston by using force balance in the vertical direction:

[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]

∴

(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05

P = 500248 Pa

P = 500.25 kPa

At state 1:

[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]

[tex]x_1 = 0.85[/tex]

Hence, this is a saturated mixture of liquid and vapor

Using the steam tables at 500.25 kPa

[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]

∴

Specific volume at state 1 is given as:

[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]

volume at state 1 is given by:

[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]

Similarly, the specific internal energy is:

[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]

[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]

[tex]U_1 = 2272.57 \ kJ/kg[/tex]

At state 2:

[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]

Using steam tables at P = 500.25 kPa and T = 320° C

[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]

∴

[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]

[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]

[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]