Answer:
No, the truck will not cross the barrier.
The closeness of the truck to the barrier is of 21.875 m
Solution:
As per the question:
Velocity of the truck, v = 25.0 m/s
Acceleration of the truck, a = - 4 
Now,
Since, the barrier at a distance of 100 m. Thus in order to check whether the truck hit the barrier or not, we will see the distance, d it covers by using the kinematic eqn:
Final velocity, v' = 0 m/s
Initial velocity = v
Now,
d = 78.125 m
Thus the truck will not cross the barrier.
Distance between the barrier and the truck:
100 - 78.125 = 21.875 m
Potential energy due to gravity = Ep = mgh [symbols have their usual meaning ]
Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.
Answer:
B = E/c = 14.04T₁ = 11 pT
Explanation:
We know c = E/B where E = maximum electric field = 3.30 × 10⁻³ V/m, B = maximum magnetic field and c = speed of light
B = E/c also c = fλ = λ/T where λ = wavelength = 235 μm = 235 × 10⁻⁶ m and T = period
c = λ₁/T₁ = λ₂/T₂ T₂ = 2.8T₁ where λ₁,λ₂ are the initial and final wavelengths and T₁,T₂ are the initial and final periods.
T₁ = λ₁/c = 235 × 10⁻⁶ m/3 × 10⁸ m/s = 7.833 × 10⁻¹³ s = 0.7833 ps
T₂ = 2.8T₁ = 2.8 × 7.833 × 10⁻¹³ s = 21.93 × 10⁻¹³ s = 2.193 ps
λ₁/T₁ = λ₂/2.8T₁
λ₂ = 2.8λ₁ = 2.8 × 235 μm = 658 μm
c = λ₂/T₂ = 2.8λ₁/2.8T₁ = λ₁/T₁ , since the speed of light c is constant.
B = E/c = E/λ₁/T₁ = ET₁/λ₁
B = ET₁/λ₁ = 3.30 × 10⁻³ V/m × T₁/235 × 10⁻⁶ m = 14.04T₁ Tesla
B = 14.04 × 7.833 × 10⁻¹³ s = 10.99 × 10⁻¹² T ≅ 11 pT
an amorphous solid is a solid that lacks the long-range order
