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Softa [21]
3 years ago
7

Melissa uses bags of mulch that way 40 pounds each. She needs to carry 10 bags to her truck. She usually carries one bag at a ti

me. Would she do less work if she carried two bags at a time?
Physics
1 answer:
arsen [322]3 years ago
4 0
No. She would be doing the same amount of work that way. Work is defined to be equal to the force multiplied by the distance. Carrying two bags at a time would cause her to exert twice the effort, so the total amount of work done in the end would still be the same.
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A sample of 5.6 L of a gas is at a pressure of 1.5 atm. If the volume of the gas is compressed to 4.8 L, what will the new press
Effectus [21]

Answer:

1.75atm

Explanation:

According to Boyle's law, the pressure P of a fixed mass of gas is inversely proportional to it's volume V provided that the temperature remains constant.

P\alpha \frac{1}{V}\\hence\\PV=constant

This implies the following;

P_1V_1=P_2V_2=...=P_nV_n............(1) Provided temperature is kept constant.

Given;

P_1=1.5atm\\V_1=5.6L\\P_2=?\\V_2=4.8L

From equation (1), we can write;

P_1V_1=P_2V_2\\hence\\1.5*5.6=P_2*4.8\\\\P_2=\frac{1.5*5.6}{4.8}\\\\P_2=1.75atm

Since all the units are consistent, there is no need for conversion.

3 0
3 years ago
Read 2 more answers
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
In the middle of a thunderstorm, a lightning bolt flashes. It takes Roberto 5 seconds to
Gemiola [76]

Answer:

 v = 344.1 m / s    

 d = 1720.5 m

Explanation:

For this problem we must calculate the speed of sound in air at 22ºC

           v = 331 RA (1+ T / 273)

we calculate

           v = 331 RA (1 + 22/273)

           v = 344.1 m / s

the speed of the wave is constant,

           v = d / t

           d = v t

we calculate

           d = 344.1   5

           d = 1720.5 m

5 0
2 years ago
Convert 123,453 to a scientific notation
Natalija [7]

Answer:

1.23453*10^5 is scientific way

3 0
3 years ago
A force is applied to an object at rest with a mass of 100 kg. A force twice as large is applied to another object at rest with
MAVERICK [17]
The second object, the one that had twice the force applied to it, would move twice as far, I believe.
3 0
3 years ago
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