Answer:
The other angle is 120°.
Explanation:
Given that,
Angle = 60
Speed = 5.0
We need to calculate the range
Using formula of range
...(I)
The range for the other angle is
....(II)
Here, distance and speed are same
On comparing both range
Hence, The other angle is 120°
Answer:
c. initial (x and y)
Explanation:
When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.
Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"
Thus, this method resolves the initial x and y velocities.
Answer:
Mass, m = 26.54kg
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
Where;
- Fapp is the applied force
- Fg is the force due to gravitation
<u>Given the following data;</u>
Net force, Fnet = 345
Acceleration, a = 3.2m/s²
<u>To find mass;</u>
Fnet = Fapp + Fg
Fnet = ma + mg
Fnet = m(a+g)
m = Fnet/(a+g)
We know that acceleration due to gravity, g = 9.8m/s²
Substituting into the equation, we have;
m = 345/(3.2 + 9.8)
m = 345/13
Mass, m = 26.54kg
<h3>
Answer:</h3>
49 N
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of the brick as 3 kg
- The coefficient of friction as 0.6
We are required to determine the force that must be applied by the woman so the brick does not fall.
- We need to importantly note that;
- For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
- That is; Friction force = Mg
But; Friction force = μ F
Therefore;
μ F = mg
0.6 F = 3 × 9.8
0.6 F = 29.4
F = 49 N
Therefore, she must use a force of 49 N
Violet light is at the end of the visible light section of the electromagnetic spectrum. Ultraviolet rays are directly next to violet rays on the EM Spectrum.
