<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
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let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
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t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
Answer:
I am pretty sure it is B My friend hope you are well
Explanation:
Answer:
The temperature is 90.4°C
Explanation:
See the attached for explanation
Answer:
W = 1,307 10⁶ J
Explanation:
Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)
F = G m₁ M / r²
W = ∫ F. dr
W = G m₁ M ∫ dr / r²
we integrate
W = G m₁ M (-1 / r)
We evaluate between the limits, lower r = R_ Moon and r = ∞
W = -G m₁ M (1 /∞ - 1 / R_moon)
W = G m1 M / r_moon
Body weight is
W = mg
m = W / g
The mass is constant, so we can find it with the initial data
For the capsule
m = 1000/32 = 165 / g_moon
g_moom = 165 32/1000
.g_moon = 5.28 ft / s²
I think it is easier to follow the exercise in SI system
W_capsule = 1000 pound (1 kg / 2.20 pounds)
W_capsule = 454 N
W = m_capsule g
m_capsule = W / g
m = 454 /9.8
m_capsule = 46,327 kg
Let's calculate
W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶
W = 1,307 10⁶ J
Answer:
Block A
Explanation:
Block A will float higher in the water compared to the second Block.
The density of water is 1g/cm³.
According to the principle of floatation "an object that floats in a liquid will displace equal amount of fluid to the weight of the object".
A body will become more submerged in water if it has more density because density is the mass per volume of body.
An object with a higher density than another will sink in the liquid of the one with lesser density.
- Object A has lesser density and will float higher up and displace very little water.
- Object B has higher density and will be more submerged.
