A 80.00 kg object is hanging from two ropes. One rope (F1) is pulling due West. The second rope (F2) is pulling 80.00 degrees up of East. Calculate the force in the first rope (F1) in Newtons. How would I do this?
1) a car moving at 50km/h crashes Into the barrier and stops in 0.25 m a) if a 10kg child were to be stopped in the same time , what is the average force must be exerted ? c)approximately what is the mass of an object whose weight equals to the force in Part b?Could you lift such a mass with your arms ? d) what does your answer to part c say about holding an infant on your lap instead of using a separate infant restraint? PS : I know answer abc but am stuck with d! A) t: 0.036s b) delta P : 139kg.m/s2 F: -3858N c) m : 393.7 no we can't ! D) ????? 2 ) If you jump off a table as your feet hit the floor let your legs bend at the knees explain why.5) a) 35g bullet Strikes at 5 kg stationary wooden block and embeds itself in the block the block and the Bullet fly out together at 8.6 m/s what was the original velocity of the bullet? 6) a 0.50kg ball traveling at 6m/s collides head on with a 1kg ball moving in the opposite direction at a velocity of -12m/s . The 0.50 ball Moves away at -14.0 m/s up aftet the collosion find the velocity of the second ball. Please THERE IS A LOT OF QUETIONS BUT THESE ARE THE ONES I DON'T GET!HELP
To calculate the force in the first rope (F1), we can break down the forces acting on the object into their horizontal and vertical components.
First, let's consider the horizontal components. The force in the first rope (F1) is pulling due West, so it is entirely in the horizontal direction. We can calculate this force using the equation:
F1_horizontal = F1 * cos(θ)
Where F1_horizontal is the horizontal component of F1, F1 is the magnitude of F1, and θ is the angle between F1 and the horizontal axis.
Since F1 is entirely in the horizontal direction, and the angle between F1 and the horizontal axis is 0 degrees, the cosine of 0 degrees is 1. Therefore, the force in the first rope (F1) is equal to its horizontal component:
F1 = F1_horizontal
Now, we need to convert the weight of the object into a force. The weight of the object is given by the equation:
Weight = mass * gravity
Where mass is the mass of the object and gravity is the acceleration due to gravity. In this case, the mass is given as 80.00 kg. To calculate the weight, we need to know the acceleration due to gravity, which is approximately 9.8 m/s^2.
Weight = 80.00 kg * 9.8 m/s^2
Next, we need to calculate the vertical component of the force in the second rope (F2). The angle between F2 and the vertical axis is (90° - 80°) = 10°. We can calculate the vertical component of F2 using a similar equation:
F2_vertical = F2 * sin(θ)
Where F2_vertical is the vertical component of F2 and θ is the angle between F2 and the vertical axis. However, we don't have the magnitude of F2 given in the question, so we need to find it.
To find the magnitude of F2, we can use the Pythagorean theorem. The object is in equilibrium, meaning the vertical forces balance out the horizontal forces. Therefore, the vertical component of F1 (F1_vertical) must be equal to the vertical component of F2 (F2_vertical). Since F1 is entirely horizontal, F1_vertical is 0. Therefore:
F2_vertical = F1_vertical = 0
Now that we know F2_vertical = 0, we can solve for the magnitude of F2 using the equation:
F2 = sqrt(F2_horizontal^2 + F2_vertical^2)
Here, F2_horizontal is the horizontal component of F2, which is equal to F2. So:
F2 = sqrt(F2^2 + 0^2) = F2
Since the vertical component is 0, the magnitude of F2 is equal to its horizontal component.
Now, we've determined that the vertical component of F2 is 0, so it does not contribute to the force that is balancing the object. Therefore, the only force acting horizontally is F1, which must be equal to the weight of the object for equilibrium.
Therefore, the force in the first rope (F1) is equal to the weight of the object:
F1 = Weight = 80.00 kg * 9.8 m/s^2
Calculate this expression to find the force in the first rope (F1) in Newtons.