Boris's reaction time is denoted as t(r), implying that he has not jumped prior to that moment. Therefore, H(b)(t) equals 0
. The vertical displacement is determined simply as
D(t) = H(a)(t)
25.82 m/s
Explanation:
Given:
Force applied by the baseball player; F = 100 N
Distance the ball travels; d = 0.5 m
Mass of the ball; m = 0.15 kg
To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved.
It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows:
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
Solving gives:
v² = (2 × 100 × 0.5) / 0.15
v² = 666.67
v = √666.67
v = 25.82 m/s.
Answer:
All observers are accurate.
Explanation:
This situation reflects a matter of reference frames regarding the book's motion as perceived by different observers.
From their distinct frames of reference, each observer's perspective is valid.
Observer A is in an inertial reference frame.
Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.
Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.
Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.
This is exemplified by observer B, who notes the car's force impacting the book's velocity.
Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.
Answer:
the temperature on the left side is 1.48 times greater than that on the right
Explanation:
GIVEN DATA:
T1 = 525 K
T2 = 275 K
It is known that
n and v are constant on both sides. Therefore we have
..............1
let the final pressure be P and the temperature 
..................2
similarly
.............3
divide equation (2) by equation (3)
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)
thus, the left side temperature equals 1.48 times the right side temperature
Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
The velocity at the stagnation point is zero.
The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Thus, the answer is 2.2 psi
