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BlackZzzverrR [31]

3 years ago

7

a trolley of mass 20kg was originally at rest on a smoth horizontal surface. bu how much will it accelerate if a pulling force o

f 22N is applied on it horizontally?

1 answer:

DochEvi [55]3 years ago

6 0

Answer:

just divide 22 N by 20 kg to get the acceleration in m/s2

Explanation:

I hope this is right-

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Hemoglobin (Hb) is the O2-carrying protein in our blood. Unlike myoglobin, it has four sites allowing it to bind up to four O2 m

bogdanovich [222]

Answer:

Yes, the energy is not simply the sum of the individual binding energies at each site, it is the product of energy at each binding site of hemoglobin.

Explanation:

Myoglobin and hemoglobin are two different cells. Myoglobin binds only one oxygen while the hemoglobin has the ability to binds four oxygen atoms at its four sides. Myoglobin present in muscle tissue only while hemoglobin is present in the whole body. Oxyhemoglobin is formed when oxygen binds with hemoglobin cell. This oxygen is take to all cells and energy is released due to the breakdown of glucose molecules with this oxygen.

4 0

4 years ago

Two narrow, parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the

CaHeK987 [17]

Answer:

a) 7.947 radians

b) $\mathbf{\frac{I}{I_{max}}=0.4535}$

Explanation:

y = Distance from central bright fringe = 2.5 mm

λ = Wavelength = 600 nm

L = Distance between screen and source = 2.8 m

d = Slit distance = 0.85 mm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{2.5}{2800}=0.000892\\\Rightarrow \theta=tan^{-1}0.000892=0.05115^{\circ}$

$\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589\times10^{-7}$

a) Phase difference

$\phi=\frac{2\pi}{\lambda}\Delta r\\\Rightarrow \phi=\frac{2\pi}{600\times 10^{-9}}7.589\times10^{-7}=7.947\ rad$

∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians

b) $\frac{I}{I_{max}}=cos^2\frac{\phi}{2}\\\Rightarrow \frac{I}{I_{max}}=cos^2\frac{7.947}{2}\\\Rightarrow \mathbf{\frac{I}{I_{max}}=0.4535}$

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe $\mathbf{\frac{I}{I_{max}}=0.4535}$

8 0

4 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

Determining wavelength

shutvik [7]

Answer:

7.5E-7

Explanation:

you would use the formula A=c/f (a= wavelength)

from here you would plug in all known values then solve.

5 0

3 years ago

Read 2 more answers

Help with 26 please:):)

Alik [6]

$0.080 per kWh

25W porch light

a day = 24hrs

365 days in a year = 8760hrs

1) convert 25W to kW

0.025kW in the porch light

2) kW in a year

0.025kW × 8760hrs = 219 kWh in a whole year

3) money

$0.080 × 219kWh = $17.52

8 0

3 years ago