Temperature is just a measure of how HOT or COLD a substance is, which can be easily defined by a magnitude using a numerical value say “300 K” or “27°C”. Hence we can say it is a scalar quantity.
But the energy which transfer by virtue of a temperature difference is a vector quantity, as it has both magnitude and direction of motion (from High temperature to low temperature region).
Answer:
The velocity of the skateboard is 0.774 m/s.
Explanation:
Given that,
The spring constant of the spring, k = 3086 N/m
The spring is stretched 18 cm or 0.18 m
Mass of the student, m = 100 kg
Potential energy of the spring, 
To find,
The velocity of the car.
Solution,
It is a case of conservation of energy. The total energy of the system remains conserved. So,
v = 0.774 m/s
So, the velocity of the skateboard is 0.774 m/s.
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Answer: An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion
Answer:
The speed of the large cart after collision is 0.301 m/s.
Explanation:
Given that,
Mass of the cart, 
Initial speed of the cart, 
Mass of the larger cart, 
Initial speed of the larger cart, 
After the collision,
Final speed of the smaller cart, 
(as its recolis)
To find,
The speed of the large cart after collision.
Solution,
Let 
is the speed of the large cart after collision. It can be calculated using conservation of momentum as :
So, the speed of the large cart after collision is 0.301 m/s.
