p=d×h×g
p=1000×10×10
:.:p=100000 pa
Explanation:
p=Liquid pressure
d=Density
h=Height
g=acceleration due to gravity
T H A N K Y O U ! !
Velocity (v) is a vector quantity that includes the magnitude and the direction. The magnitude of the quantity is the quotient obtained in dividing the distance (d) by the time (t).
v = d / t
Substituting the known values,
v = (5 km) / (25 minutes) = (5000 m) / (1500 s) =3.33 m/s
Therefore, Walt's velocity is 3.33 m/s due east.
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
a. W = 2.17 x 10 ⁴ J
b. ΔU = - 5.03 x 10⁴ J
c. Q = 2.86 x 10 ⁻³ J
Explanation:
a.
The student work does 2.17 x 10⁴ J
So W = 2.17 x 10 ⁴ J
b.
The internal energy decreases 5.03 x 10⁴ J
ΔU = U₁ - U₂
ΔU = 0 - 5.03 x 10⁴ J = - 5.03 x 10⁴ J
c.
The formula heat can use:
Q = W + ΔU
Q = 2.17 x 10⁴ J + ( - 5.03 x 10⁴ J )
Q = 2.86 x 10 ⁻³ J
