Answer:
A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.
Explanation:
The mass of copper block = 7kg
Initial speed = 4.0 m/s
Specific heat of copper = 0.385 j/g degree Celcius.
a. The increase in temperature is calculated below:
85% of energy is converted into internal energy.
b. The remaining 15 per cent of kinetic energy will be lost and it will be changed into other forms.
Time before projectile hits wall
= 88.2 m / 29.4 m/s = 3 seconds
Vertical velocity of projectile after three seconds
= 3*9.8 = 29.4 m/s
Horizontal velocity of projectile after three seconds, assuming no air resistance
= 29.4 m/s (given)
Conclusion:
velocity of projectile when it hits the wall
= < 29.4, -29.4> m/s
= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal
= 41.58 m/s east-bound at 45 degrees below horizontal.
Answer:
b.
Explanation:
Mitosis is important for growth and development of living organisms.
Answer:
44 N/m
Explanation:
The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m
The work needed to stretch a spring by <em>e</em> is given by
where <em>k</em> is spring constant.
Using the appropriate values,
Refer to the diagram shown below.
Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²
The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J
Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J
Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s
Answer: 2.324 m/s
