Answer:
the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s
Explanation:
given information
car's mass, m = 1200 kg
= 100 m
=
= 150 m
= 0
according to conservative energy
the distance from point A to B, h = 150 m - 100 m = 50 m
the initial speed
final speed = 0
thus,
² =
² - 2 g h
0 = ² - 2 g h
² = 2 g h
= √2 g h
= √2 (9.8) (50)
= 31.3 m/s
The work done on the barbell is -165.62 Nm.
Explanation:
Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.
In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.
Thus,
Now, the force is -127.4 N and the displacement is 1.3 m.
So,
So, the work done on the barbell is -165.62 Nm.
Answer:
A) x _electron = 0.66 10² m , B) x _Eart = 1.13 10² m , C) d_sphere = 1.37 10⁻² mm
Explanation:
A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of
Let's use proportions rule
x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5
x _electron = 0.66 10⁵ mm = 0.66 10² m
B) the radii of the Earth and the sun are
= 6.37 10⁶ m
tex]R_{Sum}[/tex] = 6.96 10⁸ m
Distance = 1.5 10¹¹ m
x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5
x _Eart = 1.13 10² m
C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions
d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶
d_sphere = 1.37 10⁻² mm