Answer:
40·919 m
Explanation:
Initial velocity of the arrow = 46 m/s
Angle at which it is thrown from horizontal = 38°
<h3>At the maximum height, the vertical component of velocity will be 0</h3>
Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s
From the formula
<h3>v² - u² = 2 × a × s</h3>
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Considering the formula in vertical direction and taking upward direction as positive
v = 0
u = 28·32 m/s
a = - g = - 9·8 m/s²
Let s be the maximum height
- 28·32² = - 2 × 9.8 × s
⇒ s = 40·919 m
∴ The arrow will go 40·919 m high
Force applied = F = 628 N
<span>Acceleration = a m/s² </span>
<span>Newton's 2nd law of motion : F = Ma </span>
<span> a = F/M -------- (1) </span>
<span>New mass of the crate = M1 = 3.8M kg </span>
<span>New acceleration = a1 = F/M1 = F/(3.8 M) ----- (2) </span>
<span>a1/a = {F/(3.8M)}/(F/M) = 1/3.8 = 10/38 = 5/19 ------- Answer</span>
Answer:
thermal conductivity = 0.2906 W/m.K
Explanation:
thermal conductivity = 0.168 BTU/h.ft⁰F
1 BTU/h.ft⁰F = 1.7295772056 W/m.k
0.168 BTU/h.ft⁰F = 0.168 * 1.7295772056 W/m.k
0.168 BTU/h.ft⁰F = 0.2906 W/m.k
Answer:
15.8m/s
Explanation:
This problem can be solved by taking into account the conservation of the momentum. In this case the momentum of the astronaut and the bag of tools must equal the momentum of the astronaut and the bag of tool after the astronaut throws the bag.
Hence, we have

where ma and va are the mass and velocity of the astronaut, mb and vb are the mass and velocity of the bag, after the astronaut throw the bag. The velocity v is the velocity where the astronaut has the bag of tool
By taking into account that the velocity of the astronaut must be zero to keep him near of the space station, we have that vb = 0.
Thus
