Answer:
155.25 atm
Explanation:
Distance, h = 1.6 km = 1600 m
density of water = 13.6 times less than the density of mercury
density of mercury = 13.6 x 10^3 kg/m^3
So, density of water = 1 x 10^3 kg/m^3
Pressure exerted by the water
P = distance x density of water x acceleration due to gravity
P = 1600 x 1000 x 9.8
We know that
1 atm = 1.01 x 10^5 Pa
So, =
Answer:
The capacite is C=5.32 uF using the equations of voltage and energy in capacitance
Explanation:
The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J
Using:
Voltage in this case is the energy dissipated so
Using the equation to find capacitance
F
C= 5.32 uF because u is the symbol for micro that is equal to
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
Answer:
Weight at the surface of Jupiter's moon Io is 8.13 N .
Explanation:
Given :
Acceleration due to gravity at the surface of Jupiter's moon is .
Weight of watermelon in earth , .
Acceleration due to gravity at the surface of earth is .
We know , weight is given by :
Therefore , mass at the surface of Jupiter's moon Io is :
Hence , this is the required solution .
Answer:
The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>
Explanation:
The directions in which the delivery truck travels are;
1) 2.8 km North = 2.8·, in vector form
2) 1.0 km East = 1.0·, in vector form
3) 1.6 km South = -1.6·, in vector form
Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;
Final displacement, Δd = 2.8· + 1.0·
+(-1.6·
) = 1.2·
+ 1.0·
Final displacement, = 1.0· + 1.2·
Where;
Δx = The displacement in the x-direction = 1.0·
Δy = The displacement in the y-direction = 1.2·
The magnitude of the resultant displacement vector is given as follows
= √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)
The magnitude of the resultant displacement vector ≈ 1.6 km
The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE
Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.