Answer:
1800 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
Explanation:
We know this because of Newton's first law,
, which shows us that the force on an object is equal to its mass times the acceleration it recieves. This means that taking our values of 900N and 0.5kg, and plugging them in,
![900N=(0.5kg)*(1800m/s^2)](https://tex.z-dn.net/?f=900N%3D%280.5kg%29%2A%281800m%2Fs%5E2%29)
This is honestly a little strange because the force applied and the acceleration seem ridiculous, and a little strange for an answer. Either the values are not meant to be nearly close to reality, or you made a typo.
Answer:![F_{v} =\mu_{k} mg](https://tex.z-dn.net/?f=F_%7Bv%7D%20%3D%5Cmu_%7Bk%7D%20mg)
Magnitude of the force is 2601.9 N
Explanation:
m = 450 kg
coefficient of static friction μs = 0.73
coefficient of kinetic friction is μk = 0.59
The force required to start crate moving is
.
but once crate starts moving the force of friction is reduced
.
Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie
.
Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.
Magnitude of the force
![F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N](https://tex.z-dn.net/?f=F_%7Bv%7D%20%3D%5Cmu_%7Bk%7D%20mg%5C%5CF_%7Bv%7D%20%3D0.59%20%5Ctimes%20450%20%5Ctimes%209.8%5C%5CF_%7Bv%7D%20%3D2601.9%20%20N)
So 10 gallons of gas would let you travel 300 Miles.
x gallons = 50 Miles
10 : 300 :: x : 50
x = 500/300
x = 1.66667 gallons.
So, the car would run 10 - 1.6666 gallons = 8.33 gallons.
After that, the warning light turns ON!
Hope this helps!!
Boyles law
Pressure and volume are inversely proportional as the new variable changes from the known.
Double the pressure equals 1/2 of original volume, assuming temperature remains the same.
So 40.0 mL is the new volume as it is compressed.
Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:
![45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}](https://tex.z-dn.net/?f=45%5Cfrac%7Bkm%7D%7Bh%7D%2A%5Cfrac%7B1h%7D%7B3600s%7D%2A%5Cfrac%7B1000m%7D%7B1km%7D%3D12.5%5Cfrac%7Bm%7D%7Bs%7D)
![0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s](https://tex.z-dn.net/?f=0%3D25%2B12.5t-2t%5E2%5C%5C%5C%5C2t%5E2-12.5t-25%3D0%5C%5C%5C%5Ct_%7B1%2C2%7D%3D%5Cfrac%7B-%28-12.5%29%5Cpm%20%5Csqrt%7B%28-12.5%29%5E2-4%282%29%28-25%29%7D%7D%7B2%282%29%7D%5C%5C%5C%5Ct_%7B1%2C2%7D%3D%5Cfrac%7B12.25%5Cpm%2018.87%7D%7B4%7D%5C%5C%5C%5Ct_1%3D7.78s%5C%5C%5C%5Ct_2%3D-1.65s)
You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.