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aev [14]
2 years ago
15

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop

. She drops a wooden block with a mass of 0.5 kilograms on a sensor on the floor. The sensor measures the force of the impact as 4.9 newtons. What’s the acceleration of the wooden block when it hits the sensor? Use F = ma.
Physics
2 answers:
lilavasa [31]2 years ago
6 0

Answer:

a= 9.8

Explanation:

anastassius [24]2 years ago
4 0
To calculate the acceleration of the wooden block, we use the expression F=ma where F is the force applied, m is the mass of the object and a is the acceleration. We calculate as follows:

F = ma
4.9 = 0.5a
a = 9.8 

Hope this answers the question. Have a nice day.

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Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0
2 years ago
Water (cp = 4180 J/kg·K) is to be heated by solar-heated hot air (cp = 1010 J/kg·K) in a double-pipe counter-flow heat exchanger
Inessa [10]

Answer:

452%

Explanation:

3 0
2 years ago
If a 10 meter ramp helps you move a 500 kg object up 1 meter. What was the mechanical advantage of the ramp?
sattari [20]
The mechanical advantage of an inclined plane is

(Length of the incline) / (its height)

= (10m) / (1m)

= 10 .

It's the same for any load, and doesn't depend on the mass that you're trying to move up or down the ramp.
6 0
2 years ago
Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance ove
Nuetrik [128]

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × 10^{-8} s

speed = 0.8 c = 0.8 × 3 × 10^{8}

to find out

expected distance over

solution

we know c is speed of light in air is 3 × 10^{8} m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × 10^{-8} ×  0.8 × 3 × 10^{8}

expected distance = 4.32

so the expected distance is 4.32 m

5 0
2 years ago
Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
3 0
2 years ago
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