Answer:
(a) work required to lift the object is 1029 J
(b) the gravitational potential energy gained by this object is 1029 J
Explanation:
Given;
mass of the object, m = 35 kg
height through which the object was lifted, h = 3 m
(a) work required to lift the object
W = F x d
W = (mg) x h
W = 35 x 9.8 x 3
W = 1029 J
(b) the gravitational potential energy gained by this object is calculated as;
ΔP.E = Pf - Pi
where;
Pi is the initial gravitational potential energy, at initial height (hi = 0)
ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)
ΔP.E = 1029 J
A graduated cylinder measures the volume of a liquid.
A stopwatch measures the amount of time that elapses.
A scale measures the mass of objects.
A thermometer measures the temperature of any object.
Because we are measuring rain, a liquid, we would want to use a tool that would allow us to collect the rain for measuring. Therefore, the tool e would use to measure the amount of rainfall would be A. a graduated cylinder.
Answer:
a = 1.5*10^-3 m/s^2
x = 0.033m = 3.3cm
Explanation:
To calculate the acceleration and the distance traveled by the car you use the following formulas:
(1)
(2)
v: final velocity = 0,255 km/h
vo: initial velocity = 0 m/s
t: time = 3/4 min
a: acceleration = ?
x: distance
In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.
Next, you solve the equation (1) for the acceleration a:
With this value of a you can calculate the distance traveled by the car, by using the equation (2):
hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
The pressure value is given by the equation,
Where,
represents the density of the liquid
g= gravity
h= Heigth
A) For the measurement of the guage pressure we have the data data,
Replacing we get,
![[tex]P_g = (13.6*10^3)(9.8)(0.0930)](https://tex.z-dn.net/?f=%5Btex%5DP_g%20%3D%20%2813.6%2A10%5E3%29%289.8%29%280.0930%29)
P_g = 12395Pa[/tex]
In order to find the Absolute pressure, we perform a sum between the atmospheric pressure and that of the Gauge,
B) The atmospheric pressure at sea level is 101325Pa, assuming ideal conditions, we will take this pressure for our calculation, so
