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Vladimir79 [104]
3 years ago
10

A student measures the time it takes for two reactions to be completed reaction eight is completed and 57 seconds and reaction b

e is completed and 48 seconds what can about the rights of these reactions. HELP ASAP!!
A.) The rate of reaction A is higher.
B.) The rate of reaction B is higher.
C.) The rates of reactions A and B are equal.
Physics
1 answer:
Cerrena [4.2K]3 years ago
5 0

Answer:

the rate of reaction b is higher

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A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h
BartSMP [9]
Specific Gravity of the fluid = 1.25 
Height h = 28 in
 Atmospheric Pressure = 12.7 psia
 Density of water = 62.4 lbm/ft^3 at 32F
 Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
 Difference in pressure = 1.26 psia
 (a) Pressure in the arm that is at Higher 
 P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
 (b) Pressure in the tank that is at Lower
 P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
4 0
3 years ago
A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
A particle is moving with (SHM) of period 8.0s and amplitude5.0m
nadezda [96]

Answer:

velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)

Max speed = \frac{15\, \pi}{4} \,\, \frac{m}{s}

Max acceleration = \frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}

Explanation:

Given the description of period and amplitude, the SHM could be described by:

f(x)=5\,sin(\frac{\pi}{4}x)

and its angular velocity can be calculated doing the derivative:

f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)  and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = \frac{15\, \pi}{4} \,\, \frac{m}{s}

The acceleration is found from the derivative of the velocity expression, and therefore given by:

acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = \frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}

6 0
3 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
Which segment represent slowing down
lakkis [162]

Answer:

A downward sloping line

Explanation:

6 0
3 years ago
Read 2 more answers
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