Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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We are given with
distance traveled through vacuum = 1.0 m
refractive index of water = 1.33
refractive index of glass = 1.50
refractive index of diamond = 2.42
distance traveled through water is = 1.0/1.33 = 0.75 m
distance traveled through water is = 1.0/1.50 = 0.67 m
distance traveled through water is = 1.0/2.42 = 0.41 m
Answer:
20 meters.
Explanation:
In the graph, the x-axis (the horizontal axis) represents the time, while the y-axis (the vertical axis) represents the distance.
If we want to find the distance covered in the first T seconds, you need to find the value T in the horizontal axis.
Once you find it, we draw a vertical line, in the point where this vertical line touches the graph, we now draw a horizontal line. This horizontal line will intersect the y-axis in a given value. That value is the total distance travelled by the time T.
In this case, we want to find the total distance that David ran in the first 4 seconds.
Then we need to find the value 4 seconds in the horizontal axis. Now we perform the above steps, and we will find that the correspondent y-value is 20.
This means that in the first 4 seconds, David ran a distance of 20 meters.
One km^3 is 1,000,000,000 m^3=10^9 m^3 hence 3.73 10^8 km^3 is 3.73 10^17 m^3
One meter is 3.28084 feet hence 1 m^3 is (3.28084)^3 feet
Thus 3.73 10^8 km^3 is 3.73*35.315 10^17 = 132 cubic feet
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
