Answered: During walking, a 90 kg older adult… | bartleby

Related questions

Question

During walking, a 90 kg older
adult starts at rest and begins
moving (on carpet, coefficient of
friction of 0.40) with a horizontal
velocity of 1.05 m/s. It takes them
0.25 seconds to reach this
velocity. This movement comes as
a result of a GRF that is directed
upward and forward above the
horizontal. The only other force in
the problem is gravity. How fast
are they moving vertically at the
end of this scenario?

Expert Solution

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

Two blocks are connected by a massless rope over a massless, frictionless pulley, as shown in the figure. The mass of block 2 is m2 12.1 kg, and the coefficient of kinetic friction 0.200. The angle 0 of between block 2 and the incline is µk the incline is 33.5°. If block 2 is moving up the incline at 2 constant speed, what is the mass mj of block 1? 1 kg M¡ =
A skateboard with mass of 50kg is pulled without friction across the ground by a rope that makes an angle of 20 degrees with the horizontal. What force on the rope would be required to move the skateboard at a constant speed of 20 m/s?
Determine the force Q-> when the block moves with constant velocity. Express your answer in vector form.
A child riding a bicycle has a total mass of 30.6 kg. The child approaches the top of a hill that is 27.9 m high 27.9 and is 86.8 m (ℓ) long at 12.1 m/s. If the force of friction between the bicycle and the hill is 26.7 N, what is the child's velocity at the bottom of the hill? (Disregard air resistance, g = 9.8 m/s2.)
Suppose the 50 turn coil in the figure below lies in the plane of the page and originally has an area of 0.230 m². It is stretched to have no area in 0.100 s. X Bin XBin X V What is the direction? clockwise X counterclockwise X X What is the magnitude (in V) of the average value of the induced emf if the uniform magnetic field points into the page and has a strength of 1.75 T? X X 192 X
You place a crate of mass 47.7 kg on a frictionless 3.61-meter-long incline. You release the crate from rest, and it begins to slide down, eventually reaching the bottom 1.15 s after you released it. What is the angle of the incline? 1 50.7 degrees 76.1 degrees 3. 42.3 degrees 4 33.8 degrees
Asap please.....
A woman pulls a cloth out from under a 200 g bowl sitting on a table. The bowl is 30.0 cm from the far edge of the cloth, and the cloth exerts a friction force of 0.125 N on the bowl. Assume the cloth is pulled with a constant horizontal acceleration of 2.80 m/s2. What distance (in m) does the bowl travel relative to the top of the table? (Assume the cloth moves in the +x-direction and notice that it moves more than 30.0 cm relative to the top of the table.)
A 5-kg block is set into motion up an inclined plane (30°) with an initial speed of 8.0 m/s. There is friction between the block and the plane. The block comes to rest after traveling 3.0 m along the plane. Define the positive x-direction to be down the ramp.
A woman at an airport is rolling her suitcase (mass = 20 kg) at an initial velocity of 1 m/s by pulling on a strap at an angle of 30° above the horizontal with a force of 33 N. The woman starts to speed up when she realizes that she is late for her flight. She pulls the suitcase over a distance of 50 m as she walks to her gate. What is the speed of the suitcase when she reaches her gate? (You can neglect friction in this problem. Give your answer in m/s.)
Two crates of fruit are released from the top of a ramp inclined at 30 degrees from the horizontal and 4.5 meter long. The two crates consist of an apple crate of mass 20 kg that is placed in front of a watermelon crate of mass 80 kg. The apple crate has a coefficient of friction of 0.20 while the watermelon crate has a coefficient of friction of 0.15. How long does it take the apple crate to reach the bottom of the incline if it needs to travel a distance of 4.5 meters?
A parachutist whose mass is 70 kg drops from a helicopter hovering 2500 m above the ground and falls toward the ground under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b₁ = 30 N-sec/m when the chute is closed and b₂ = 90 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches 15 m/sec, after how many seconds will the parachutist reach the ground? Assume that the acceleration due to gravity is 9.81 m/sec². - The parachutist will reach the ground after (Round to two decimal places as needed.) * seconds.

SEE MORE QUESTIONS