Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and 
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
The range is 35.35 m
I think its 13...........
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
Chalecos no tienen mangas. Vests don't have sleeves.<span />
Answer:
F₃ = 122.88 N
θ₃ = 20.63°
Explanation:
First we find the components of F₁:
For x-component:
F₁ₓ = F₁ Cos θ₁
F₁ₓ = (50 N) Cos 60°
F₁ₓ = 25 N
For y-component:
F₁y = F₁ Sin θ₁
F₁y = (50 N) Sin 60°
F₁y = 43.3 N
Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:
F₂ₓ = F₂ = 90 N
F₂y = 0 N
Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:
F₁ₓ + F₂ₓ + F₃ₓ = 0 N
25 N + 90 N + F₃ₓ = 0 N
F₃ₓ = - 115 N
for y-components:
F₁y + F₂y + F₃y = 0 N
43.3 N + 0 N + F₃y = 0 N
F₃y = - 43.3 N
Now, the magnitude of F₃ can be found as:
F₃ = √F₃ₓ² + F₃y²
F₃ = √[(- 115 N)² + (- 43.3 N)²]
<u>F₃ = 122.88 N</u>
and the direction is given as:
θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)
<u>θ₃ = 20.63°</u>
