<u>Answer</u>
27.7
<u>Explanation</u>
The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.
cosx = adjacent/hypotenuse
cos 30 = adjacent / 10
adjacent = 10 cos30
= 8.66 m/s ⇒ This is the horizontal speed.
Now find the horizontal distance.
Distance = speed × time
= 8.66 × 3.2
= 27.71
Answer to the nearest tenth = 27.7
Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
Answer: Both Technician A and B
Explanation:
There is a similar process in using a pressure transducer and lab scope to using a vacuum gauge.
And also, the pressure transducer can be used to tie any issues to individual cylinders if paired with a second trace consisting of the ignition pattern. Therefore, both Technician A and B are correct.
Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
