Questions LLC
Login
or
Sign Up
Ask a New Question
Questions
Science
john is on a skiing holiday. he starts from rest on a long gentle incline of 6° and the coefficient of kinetic friction between the skies and the snow is 0,06. the weight of john and his skies is 800 n. Traveling 100m
1 answer
Incomplete.
You can
ask a new question
or
answer this question
.
Similar Questions
john is on a skiing holiday. he starts from rest on a long gentle incline of 6° and the coefficient of kinetic friction between
Top answer:
Incomplete.
Read more.
A 31.00 kg block starts from rest at the top of a 24.0 m long 30.0° incline. Its kinetic energy at the bottom of the incline is
Top answer:
work done by friction= mgh - finalKE
Read more.
A ski starts from rest and slides doen a 22 degree incline 75m long. a) if the coefficient of friction is 0.090, what is the
Top answer:
a)20.7m/s b)242.9m
Read more.
A box slides down a frictionless 35 degree incline. Determine its acceleration. If the incline is 10.0 m long and the box starts
Top answer:
To determine the acceleration of the box sliding down the incline, we can use the following
Read more.
A 1.7 kg otter starts from rest at the top of a muddy incline 85.3 cm long and slides down to the bottom in 0.40 s.
What net
Top answer:
14.523
Read more.
A 1.5 kg otter starts from rest at the top of a muddy incline 87.1 cm long and slides down to the bottom in 0.60 s.
What net
Top answer:
12
Read more.
A 2.2 kg otter starts from rest at the top of a muddy incline 93.9 cm long and slides down to the bottom in 0.60 s.
What net
Top answer:
gravity acts on the otter, and friction acts on the otter.
Read more.
A 1.9 kg otter starts from rest at the top of a
muddy incline 90 cm long and slides down to the bottom in 0.50 s. What net
Top answer:
d = 0.5at^2 = 0.90m, 0.5a(0.5)^2 = 0.90, 0.125a = 0.90, a = 0.90 / 0.125 = 7.2m/s^2. F = ma = 1.9 *
Read more.
A 2.0Kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in .50s. What net force
Top answer:
The force (F) equals the mass (M= 2.0 kg) times the acceleration. If it travels 0.85 m in t = 0.50
Read more.
A 2 kg otter starts from rest at the top of a muddy incline 85 cm long ans slides down to the bottom in 0.5 s. What net external
Top answer:
I come up with 13.6 N as my net force. Is this correct?
Read more.
Related Questions
A 2.0 kg mass starts from rest and slides down an inclined plane 8.0 X 10^-1 m long in .50s. what net force is acting on the
A 20. kg block is placed at the top of a 10. meter-long inclined plane. The block starts from rest and slides without friction
A 3.00-kg block starts from rest at the top of a 25.5° incline and slides 2.00 m down the incline in 1.75 s.
(a) Find the
A skier is gliding along at 2.5m/s on horizontal, frictionless snow. He suddenly starts down a 10∘ incline. His speed at the
"A 25-N crate is held at rest on a frictionless incline by a force that is parallel to the incline. If the incline is 25◦
Trent is skiing on a circular ski trail that has a radius of 0.9 km. Trent starts at the 3-o'clock position and travels 2.6 km
A block slides down a frictionless plane having an inclination of θ = 13.2°. The block starts from rest at the top, and the
Car A is 12 m ahead of car B on a straight stretch of road. Car A starts from rest and accelerates away from B at 4.0 m/s2. Two
A 1.9 kg otter starts from rest at the top of a muddy incline 84.1 cm long and slides down to the bottom in 0.60 s.
What net
A 1.6 kg otter starts from rest at the top of a
muddy incline 95.9 cm long and slides down to the bottom in 0.50 s. What net