Answers
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that the magnitude of the resultant force is
R=200N
From the question we are told
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Generally the equation for the Resultant force is mathematically given as
For x axis resolution
[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]
For y axis resolution
[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]
Therefore
[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]
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Related Questions
A body moves due north with velocity 40 m/s. A force is applied
on it and the body continues to move due north with velocity 35 m/s. W. .What is the direction of rate of change of momentum,if it takes
some time for that change and what is the direction of applied
external force?
Answers
Answer:
the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
Explanation:
The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:
[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]
p: momentum
m: mass
[tex]\Delta v[/tex]: change in the velocity
The sign of the change in the velocity determines the direction of rate of change. Then you have:
[tex]\Delta v=v_2-v_1[/tex]
v2: final velocity = 35m/s
v1: initial velocity = 40m/s
[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]
Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.
The applied force is also against the direction of motion of the body, downward.
How far does a roller coaster travel if it accelerates at 2.83 m/s2 from an initial
velocity of 3.19 m/s for 12.0 s?
Answers
Answer:
b
Explanation:
A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop
Answers
Answer:
The velocity is [tex]v = 8.85 m/s[/tex]
Explanation:
From the question we are told that
The mass of the skier is [tex]m_s = 60 \ kg[/tex]
The initial speed is [tex]u = 4.0 \ m/s[/tex]
The height is [tex]h = 10 \ m[/tex]
According to the law of energy conservation
[tex]PE_t + KE_t = KE_b + PE_b[/tex]
Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as
[tex]PE_t = mg h[/tex]
substituting values
[tex]PE_t = 60 * 4*9.8[/tex]
[tex]PE_t = 2352 \ J[/tex]
And [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill
And [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as
[tex]KE_b = 0.5 * m * v^2[/tex]
substituting values
[tex]KE_b = 0.5 * 60 * v^2[/tex]
=> [tex]KE_b = 30 v^2[/tex]
Where v is the velocity at the bottom
And [tex]PE_b[/tex] is the potential energy at the bottom which equal to zero due to the fact that height is zero at the bottom of the hill
So
[tex]30 v^2 = 2352[/tex]
=> [tex]v^2 = \frac{2352}{30}[/tex]
=> [tex]v = \sqrt{ \frac{2352}{30}}[/tex]
[tex]v = 8.85 m/s[/tex]
Answer:
The Skier's velocity at the bottom of the hill will be 18m/s
Explanation:
This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.
That is
[tex]mgh = 0.5 mv^{2}[/tex]
solving for velocity, we will have
[tex]v= \sqrt{2gh}[/tex]
hence her velocity will be
[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]
This is the velocity she gains from the slope.
Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s
The Skier's velocity at the bottom of the hill will be 18m/s