Answer:
69°
Explanation:
Projectiles that land at the same elevation they're launched from will have the same range if the launch angles are complementary (add up to 90°).
The complement of 21° is 69°.
In order to find the angle of the refracted ray, we may use the Snell's law (also known as Snell–Descartes law and the law of refraction). This law states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media such as water, glass, or air.
In mathematical form, this is:
n₁sin(∅₁) = n₂sin(∅₂)
Where:n is the refractive index.
Plugging in the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91
The angle of refraction is 15°.
Answer:
12.6 m/s
Explanation:
We know that the linear speed v = rω where r = radius and ω = angular acceleration. Given that ω = 62.8 rad/s and r = 20 cm from center = 0.2 cm
So, v = rω
= 0.2 m × 62.8 rad/s
= 12.56 m/s
≅12.6 m/s
Answer:
a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s
Explanation:
In this exercise the change in moment of a ball is asked in two different cases
a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero
Δp = p_f - p₀
Δp = 0 - m v₀
Δp = - 0.100 20
Δp = -2.0 kgm / s
b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count
v_f = - v₀
Δp = p_f -p₀
Δp = m v_f - m v₀
Δp = m (v_f -v₀)
Δp = 0.100 (-20 - 20)
Δp = -4 kg m / s
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
(1)
xo = 10 km
v1 = 55km/h
car 2:
(2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
The position in which both cars coincides is:
