Answer:
This is an example of Inelastic colission
Explanation:
Step one:
given:
mass of moose m1 = 620 kg
mass of train m2= 10,000kg
Initial velocity of moose u1= 0 m/s
Initial velocity of train v1 = 10m/s
combined velocity of the system is given as v
Applying the conservation of momentum equation we have
m1u1+ m2u1= (m1+m2)V
substitutting we have
620*0+10000*10= (620+10000)V
100000= 10620V
divide both sides by 10620
V = 100000/10620
V=9.41m/s
The velocity of the moose after impact is 9.41m/s
Answer:
The forces experienced by the middle particle are attractive, and the net force will remain the same (0) if and only if the distances of the sides particles to the middle particle are the same.
Explanation:
In example 20.3 the forces experienced by the middle particle are repulsive because all the particles are positive, for the case in which the particles on the sides are replaced for negative charge particles the forces experienced by the middle particle are attractive. Regarding the net force, because we don't know the distances we can not give a definitive answer, what we can say is that if the distances from the middle particle to the sides particles are the same the net force is zero for both cases (remain unchanged).
It is FALSE.
The absolute value of both numbers is 7, therefore they are equal!
Hope this helps!
Answer:
The number of beats is 10.58 in 35°C.
Explanation:
The beat frequency is given by : f₁-f₂
At 5°C, f₁-f₂ = 4
We need to find the number of beats in 35°C.
The frequency in a standing wave is proportional to 
.
So,
So, the number of beats is 10.58 in 35°C.
Answer:
1000.66m
Explanation:
L1=1000m
Temperature 1=-20
L2=?
Temperature 2=40
Temperature difference=40-(-20)
40+20=60
inserting into the formula
l2=l1(1+α×changeintemperature)
L2=1000(1+11×10^-6 ×60)
L2=1000(1+6.6×10^-4)
L2=1000(1.000.66)
L2=1000.66m
