Answer:
the extension recorded by the student would be smaller than the actual extension of the spring
Answer:
1.
109.6 cm , - 1.74 , real
2.
1.5
Explanation:
1.
d₀ = object distance = 63 cm
f = focal length of the lens = 40 cm
d = image distance = ?
using the lens equation
d = 109.6 cm
magnification is given as
m = - 1.74
The image is real
2
d₀ = object distance = a
d = image distance = - (a + 5)
f = focal length of lens = 30 cm
using the lens equation
a = 10
magnification is given as
m = 1.5
Answer:
(a) 7.72×10⁵ J
(b) 4000 J
(c) 1.82×10⁻¹⁶ J
Explanation:
Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass, v = velocity
(a)
For a moving automobile,
Ek = 1/2mv².
Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s
Substitute into equation 1
Ek = 1/2(2.0×10³)(27.78²)
Ek = 7.72×10⁵ J
(b)
For a sprinting runner,
Given: m = 80 kg, v = 10 m/s
Substitute into equation 1 above,
Ek = 1/2(80)(10²)
Ek = 40(100)
Ek = 4000 J
(c)
For a moving electron,
Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s
Substitute into equation 1 above,
Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²
Ek = 1.82×10⁻¹⁶ J
Answer:
Nucleus And electron cloud
Explanation:
Hope this helps
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, 
Area of cross section, 
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
So, the drag coefficient is 1.01.
