Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N
Answer:
Moment of inertia = 0.3862kg-m²
Explanation:
2.00x10³
2.80cm
145 rad
r = r⊥ x F
F is an applied force
r⊥ is the distance between the applied force and axis
Force exerted = 2.00x10³
r⊥ = 2.8cm = 0.028m
Alpha = 145rad/s²
r = 0.028m x 2.00x10³
r = 56.0N-m
To get the moment of inertia
56.0N-m² = (145rad/s²) x I
The I would be:
I = (56.0N-m²)/(145rad/s²)
I = 56/145
= 0.3862Kg-m²
This is the moment of inertia.
Thank you!
By definition, the potential energy is:
U = qV
Where,
q: load
V: voltage.
Then, the kinetic energy is:
K = mv ^ 2/2
Where,
m: mass
v: speed.
As the power energy is converted into kinetic energy, we have then:
U = K
Equating equations:
qV = mv ^ 2/2
From here, we clear the speed:
v = root (2qV / m)
Substituting values we have:
v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
v = 3.56 × 10 ^ 7 m / s
Then, the centripetal force is:
Fc = Fm
mv ^ 2 / r = qvB
By clearing the magnetic field we have:
B = mv / qr
Substituting values:
B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
B = 3.43 × 10 ^ -3 T
Answer:
A magnetic field that must be experienced by the electron is:
B = 3.43 × 10 ^ -3 T
