We know the formulas for momentum and energy. But they both involve the mass of
the object, and we don't know the mass of the baseball. What can we do ?
It's not a catastrophe. The question only asks which one is bigger. If we're clever,
we can answer that without ever knowing how much the momentum or the energy
actually is. We know that both baseballs have the same mass, so let's just call it
' M ' and not worry about what it really is.
<u>Momentum of anything = (mass) x (speed)</u>
Momentum of the first baseball = (M) x (4 m/s) = 4M
Momentum of the second one = (M) x (16 m/s) = 16M
The second baseball has 4 times as much momentum as the first one has.
<u>Kinetic energy of anything = 1/2 (mass) x (speed squared)</u>
KE of the first baseball = 1/2 (M) x (4 squared) = 8M
KE of the second one = 1/2 (M) x (16 squared) = 128M
The second baseball has 16 times as much kinetic energy as the first one has.
Answer:
a) 3.37 x 
b) 6.42kg/
Explanation:
a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .
Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x 
. So density of metal = mass of metal / volume of metal = 5 / 14 x = 3.37 x
b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/
Answer:
A morse code alphabet decoder maybe?? I am confused
Explanation:
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s
Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field
k = 9 x 10⁹ N.m²/C²
q is the charge of the object
now,
q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC
