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2 years ago

12

When Aditya pushes on Rachel and her bicycle, they accelerate at 0.22 m/s/s. If Aditya pushes on Rachel and her bicycle with twi

ce as much force, then her
acceleration will be?

A. 1/4 as much
B. 1/2 as much
C. twice as much
D. the same

1 answer:

lord [1]2 years ago

4 0

I think it’s D but I’m not sure

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Which factors could be potential sources of error in the experiment? check all that apply.

Vadim26 [7]

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

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At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta

Lubov Fominskaja [6]

Answer:

a=2330

b= 0.223secs

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2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

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1 year ago

Hot combustion gases enter the nozzle of a turbojet engine at 260 kpa, 747oc, and 80 m/s. the gases exit at a pressure of 85 kpa

Aleksandr-060686 [28]

Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temp are to be determined.

Given:

T1 = 1020 K à h1 = 1068.89 kJ/kg, Pr1 = 123.4

P1 = 260 kPa

T1 = 747 degrees Celsius

V1 = 80 m/s ->nN = 92% -> P2 = 85 kPa

Solution:

From the isentropic relation,

Pr2<span> = (P2 / P1)PR1 = (85 kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>

There is only one inlet and one exit, and thus, m1 = m2 = m3. We take the nozzle as the system, which is a control volume since mass crosses the boundary.

h2a = 1068.89 kJ/kg – (((728.2 m/s)2 – (80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) = 806.95 kJ/kg\

From the air table, we read T2a = 786.3 K

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1 year ago

A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor

Alex17521 [72]

Answer:

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From the relation

PV = nRT

P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0

2 years ago