Result: 168N
The calculation shows 16 - 10 equals 6
and 6 divided by 10 equals 0.6
. Therefore, F equals 280 multiplied by 0.6 equals 168.
Response:
(a) 104 N
(b) 52 N
Clarification:
Provided Information
Incline angle of the ramp: 20°
F forms a 30° angle with the ramp
The parallel component of F along the ramp is Fx = 90 N.
The perpendicular component of F is Fy.
(a)
Consider the +x direction pointing up the slope, and the +y direction perpendicular to the ramp's surface.
Using the Pythagorean theorem, decompose F into its x-component:
Fx=Fcos30°
To find F:
F= Fx/cos30°
Insert the value for Fx based on the given info:
Fx=90 N/cos30°
=104 N
(b) Calculate the y-component of r using the Pythagorean theorem:
Fy = Fsin 30°
Substituting for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
Answer:
Please refer to the explanation
Explanation:
Race distance is 5km
Top speed = 45 yards
Converting yards to kilometers:
1km equals 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x indicates the maximum distance he can sustain his highest speed in kilometers.
Thus, from the data available, we can determine that Lamar will not be able to maintain his maximum speed for the full 5km race, as he can only sustain it for 0.0411 kilometers.
The greatest mass that can hang without submerging is 2.93 kg. The provided details are as follows: sphere diameter = 20 cm, hence the radius r = 10 cm = 0.10 m. The density of the Styrofoam sphere is 300 kg/m³. The sphere's volume calculates to 4.18 * 10⁻³ m³. Mass M = Density * Volume results in (300)(4.18 * 10⁻³ m³) = 1.25 kg. The displaced water mass is computed as volume * water density, yielding 4.18 * 10⁻³ m³ * 1000 = 4.18 kg. The additional mass the sphere can hold is the difference between the two mass calculations: 4.18 kg - 1.25 kg = 2.93 kg.
In this scenario, there exists a constant electric field produced by a large sheet. This electric field can be defined as... The force acting on the ball due to this field acts horizontally, and this force must be counterbalanced by the horizontal tension component of the string to maintain equilibrium. Similarly, the vertical tension component in the string must equal the weight of the small sphere. Hence, we can derive two equations to illustrate this.
