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kherson [118]
3 years ago
14

An 80 kg object has a µk = 0.35 and a µs = 0.60. Assuming it is on a flat surface

Physics
1 answer:
Over [174]3 years ago
6 0

Answer:

What is the normal force on the object (draw a diagram if needed)

784 N

How much force is required to get the object to start to move from rest (max static friction)?

  470.4 N

When the object starts moving, what is the force of kinetic friction?

274.4 N

If the moving object has a tension force of 300N to the right pulling it, what is the net force on the object in the horizontal direction? 784 N

What is the acceleration (with direction) of the object based on your answer for part d?  Remember that a = net force/m.

Explanation:

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A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs
defon

Answer:

2940.1 joules  would you burn in climbing stairs all day.

Explanation:

Work = W = F\times d

going up stairs  would be against force of gravity

W = mgh

where h is the  height

the question is not complete because we need speed or distance

h =  v \times t

so assuming 1 step per second

h = 86,400 steps  \times 7inchs/step \times 0.0254 m/inch

h = 15362 m

so from this    

W = 800 N \times 15362

   = 12289600 J  

that means YOU  need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal  

Kcal =  \frac{W}{(J/Kcal)}

       = \frac{12289600}{4180}

       = 2940.1 Kcal

if you ran faster you would use more energy  2 steps per second would mean 5880 Kcal.

8 0
4 years ago
XERCISE
kkurt [141]

The answer is Anguer...

<em>Hope </em><em>it </em><em>helps.</em><em>.</em><em>.</em><em> </em><em>pls </em><em>mark</em><em> brainliest</em>

3 0
3 years ago
What is the approximate weight of a 20-kg cannonball on earth?
disa [49]
The weight of an object is taken to be the force on the object due to gravity. The weight ( W ) is the product of the mass ( m ) of the object and the magnitude of the gravitational acceleration ( g ).
On Earth: g = 9.81 m/s²
m = 20 kg
W = m · g = 20 kg · 9.81 m/s² = 196.2 N
6 0
3 years ago
Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
julsineya [31]

Answer:

U=50.96J

Explanation:

The electrostatic potential energy for pair of charge is given by

U=1/4π∈₀×(q₁q₂/r)

Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:

U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

U=3×1/4π∈₀×(q²/r)

Substitute given values

So

U=3\frac{1}{4\pi E_{o} }\frac{q^{2} }{r}\\  U=3\frac{1}{4\pi8.85*10^{-12} }\frac{(1.15*10^{-6}C )^{2} }{0.0007m}\\  U=50.96J

8 0
3 years ago
An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be
dedylja [7]

Answer:

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

Explanation:

When an object experiences acceleration to the left, the net force acting on this object will also be to the left.

From Newton's second law of motion, the acceleration of the object is given as;

a = ∑F / m

a = -F / m

The negative value of "a" indicates acceleration to the left

where;

∑F is the net force on the object

m is the mass of the object

At a constant force, F = ma ⇒ m₁a₁ = m₂a₂

If the mass of the object was doubled, m₂ = 2m₁

a₂ = (m₁a₁) / (m₂)

a₂ = (m₁a₁) / (2m₁)

a₂ = ¹/₂(a₁)

Therefore, the following can be deduced from the acceleration of this object;

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

6 0
3 years ago
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