







<h3>☯ <u>By using formula of Lens</u> </h3>











<h3>☯ <u>Now, Finding the magnification </u></h3>





<h3>☯ <u>Hence</u>,

</h3>


Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration


State 2 : constant speed

State 3 : deceleration


Total distance : state 1+ state 2+state 3

the average velocity = total distance : total time

Answer:
b. Friction decreased when he went from pavement to ice and then increased two more times.
Explanation:
Frictional force depends on the normal force of the surface and a friction coefficient.

Since we're talking about the same car, the value of
will remain constant whereas μ will represent the change in the frictional coefficient of the surface. Now we consider the different surfaces, cars will slide in an icy road which means that the frictional coefficient is smaller than the pavement.
After Joshua returns to the pavement road, the resulting frictional force increases and will do so one more time when he reaches the gravel road. Gravel roads have greater frictional coefficients than pavement roads which means the frictional force will increase a second time.
Answer:
(a) m = 1.6 x 10²¹ kg
(b) K.E = 2.536 x 10¹¹ J
(c) v = 7.12 x 10⁵ m/s
Explanation:
(a)
First we find the volume of the continent:
V = L*W*H
where,
V = Volume of Slab = ?
L = Length of Slab = 4450 km = 4.45 x 10⁶ m
W = Width of Slab = 4450 km = 4.45 x 10⁶ m
H = Height of Slab = 31 km = 3.1 x 10⁴ m
Therefore,
V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)
V = 6.138 x 10¹⁷ m³
Now, we find the mass:
m = density*V
m = (2620 kg/m³)(6.138 x 10¹⁷ m³)
<u>m = 1.6 x 10²¹ kg</u>
<u></u>
(b)
The kinetic energy will be:
K.E = (1/2)mv²
where,
v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)
v = 3.17 x 10⁻¹⁰ m/s
Therefore,
K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²
<u>K.E = 2.536 x 10¹¹ J</u>
<u></u>
(c)
For the same kinetic energy but mass = 77 kg:
K.E = (1/2)mv²
2.536 x 10¹¹ J = (1/2)(77 kg)v²
v = √(2)(2.536 x 10¹¹ J)
<u>v = 7.12 x 10⁵ m/s</u>