Which factors can change the size of the shadow of an opaque object
2 answers:
Answer: c)how far the object is from the light source
d)the size of the screen
The size of the shadow depends on the distance of the source of light and on the angle at which the light rays fall on the object. If the source of light is closer to the object, a larger shadow is formed than when the source of light is far from the object.
The angle at which a light strikes an object affects the size and shape of its shadow. Shadows are longer when the light source is at a low angle (side on) and shorter when the light source is at a higher angle (overhead).
The position of the light source can change the size and shape of a shadow. A good example would be, when we are outside on a sunny day, we can see how our shadows change throughout the day. The Sun's position in the sky affects the length of the shadow.
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Answer:
Explanation:
Note that acceleration is the rate change of velocity i.e
.
Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e
if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.