Answer:
Vf= 7.29 m/s
Explanation:
Two force act on the object:
1) Gravity
2) Air resistance
Upward motion:
Initial velocity = Vi= 10 m/s
Final velocity = Vf= 0 m/s
Gravity acting downward = g = -9.8 m/s²
Air resistance acting downward = a₁ = - 3 m/s²
Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²
( Acceleration is consider negative if it is in opposite direction of velocity )
Now
2as = Vf² - Vi²
⇒ 2 * (-12.8) *s = 0 - 10²
⇒-25.6 *s = -100
⇒ s = 100/ 25.6
⇒ s = 3.9 m
Downward motion:
Vi= 0 m/s
s = 3.9 m
Gravity acting downward = g = 9.8 m/s²
Air resistance acting upward = a₁ = - 3 m/s²
Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²
Now
2as = Vf² - Vi²
⇒ 2 * 6.8 * 3.9 = Vf² - 0
⇒ Vf² = 53. 125
⇒ Vf= 7.29 m/s
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀
Answer:

Explanation:
For answer this we will use the law of the conservation of the angular momentum.

so:

where
is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:



Finally we replace all the data:

Solving for
:

A. True
if cold air is replacing warm air it is a cold front and vice versa.
Answer:
1904 m
Explanation:
Here we need to find out the depth of the canyon . When you will scream after seeing a snake , the sound produced will travel till the end of the canyon and after hitting the end , it will travel back to you .
- So if the depth of the canyon is d (say) , then the total distance travelled by the sound wave will be d + d = 2d .
And we know that the speed of the sound is approximately 340m/s in air , so we can use the formula distance = speed * time to calculate the depth of the canyon . So ,
