D) Gamma Rays
Explanation:
Gamma rays can transmit through matter.
<span>What you need to do while answering this questions, is ask yourself what has cells - only if a thing has cells can you see those cells under a microscope. Objects of animal and plant origin have cells, so blood, plant and cork (made of tree bark) can have cells, and a box too, if it's made of wood. So we can''t exclude any answers based on this. We must then know the story of Robert Hook - and it was in fact a cork. He did this discovery around 1655. At the time his main interest was the microscope rather than the cork, and he used to cork to demonstrate the function of the microscope. The correct answer is CORK.</span>
Answer:
a) 45 s , b) vₐ = 90 m / s, v_b = 162 m / s, c) x_b = 3.328 10⁴ m
Explanation:
We can solve this exercise using the kinematic relations
Vehicle A
xₐ = v₀ₐ t + ½ aₐ t²
vehicle B
starts two seconds later
x_b = v_{ob} (t-2) + ½ a_b (t-2) ²
as cars start from rest their initial velocities are zero
at the point where they meet, the position must be the same for both vehicles
xa = 0 + ½ aₐ t²
xb = 0 + ½ a_b (t-2) ²
½ aₐ t² = ½ a_b (t-2) ²
t = (t-2)
t (1 - \sqrt{ \frac{a_a}{a_b} }) = 2
t (1 - ⅔, ) = 2
t = 2 / 0.4444
t = 45 s
b)
the speed of each car
vₐ = voa + aa t
vₐ = 0 + 2 45
vₐ = 90 m / s
v_b = 3.6 45
v_b = 162 m / s
c) xb = 0 + ½ ab (t-2) ²
x_b = ½ 3.6 (45-2) ²
x_b = 3.328 10⁴ m
Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
- W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
= K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
Marbles would have the greater density. Marbles would have the greater density because the molecules inside are more tightly packed. Even though sand has the greater mass it does not have the greater density. The sand doesn't have the greater density because it could be crushed in an instant, showing that if molecules are packed together tightly they will be tougher to break. The dice, I believe, still have a greater density than the sand.