The victim's head is accelerated faster and harder than the
torso when the victom is involved in a typical rear-end collision.
The traffic accident where a vehicle crashes into another
vehicle that is directly in front of it is called a rear-end collision.
One of the most common accident in the United States is the
rear-end collision, and in a lot of cases, rear-end collisions are prompted by
drivers who are inattentive, unfavorable conditions of the road, and poor
following distance.
<span>An enough room in front of your car so you can stop when the
car in front of you stops suddenly is one basic driving rule. The person isn’t
driving safely if he / she is behind you and couldn’t stop.</span>
Sound waves travel around the boxed room causing them to bounce off the nearest walls to the end of the room
Answer:
11.95m/s
Explanation:
A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.
a) Find the speed of the object. Answer in units of m
K. E =½mv²
150= ½mv²
Multiply both sides by 2
mv² = 300
Divide both sides by v²
m = 300/v² .................. Equation 1
Momentum is the product of mass and velocity
Momentum = mv
25.1 = mv
Divide both sides by v
m = 25.1/v ................ Equation 2
Equate equations 1 and 2
300/v² = 25.1/v
Cross multiply
25.1v² = 300v
Multiply v with both sides
25.1v = 300
Divide both sides by 25.1
v = 300/25.1
V = 11.95m/s
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The kinetic energy for a large vehicle is different from that of a smaller vehicle, assuming that the vehicles are travelling at the same speed and stopping the same distance. This is because for a larger vehicle the kinetic energy is higher, as the mass for a larger vehicle, is more than the smaller vehicle.
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
