Answer:
Without food your not able to produce energy thats why you cannot work so long
Answer:
The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change
Explanation:
Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.
Answer:twice of initial value
Explanation:
Given
spring compresses 
distance for some initial speed
Suppose v is the initial speed and k be the spring constant
Applying conservation of energy
kinetic energy converted into spring Elastic potential energy
When speed doubles
divide 1 and 2
Therefore spring compresses twice the initial value
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
