A ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
a = - 6.5 : 1.2
a = - 5.4167 m/s²
F = m * a = 950 kg * 5.4167 m/s²
F = 5,145.8 N ( the average force exerted on a car during braking )
b ) d = v o - a t² / 2
d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
= 19.20 - 3.90 = 15.30 m
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα= 
Senα= 
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
Answer:
F1 = G m1 m2 / R^2 force of attraction
F2 = G m1 m2 / (R/2)^2
F2 / F1 = 4 the force of gravity will be quadrupled
Answer:
a) 578.0 cm²
b) 25.18 km
Explanation:
We're given the density and mass, so first calculate the volume.
D = M / V
V = M / D
V = (6.740 g) / (19.32 g/cm³)
V = 0.3489 cm³
a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.
V = Ah
A = V / h
A = (0.3489 cm³) / (6.036×10⁻⁴ cm)
A = 578.0 cm²
b) The volume of a cylinder is pi times the square of the radius times the length.
V = πr²h
h = V / (πr²)
h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)
h = 2.518×10⁶ cm
h = 25.18 km
Answer:
joules
joules
Explanation:
Let us convert the time in hours into seconds

Change in internal energy

where E is the internal energy in Joules
p is the power in watts
and t is the time in seconds

Joules
Amount of work done by the system

where P is the pressure and V is the volume
Substituting the given values in above equation, we get -

liter-atmospheres
Work done in Joules

Joules

Substituting the given values we get -

Thus
joules
joules