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Trava [24]
3 years ago
14

As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i

s 100) A) static friction. B) kinetic friction. C) a combination of static and kinetic friction. D) neither static nor kinetic friction, but some other type of friction. E) It is impossible to tell what type of friction acts in this situation.
Physics
2 answers:
Verdich [7]3 years ago
7 0

Answer:

B) kinetic friction.

Explanation:

Remember that static friction is only the friction that is impeding the object from moving and in that matter, static friction is the amount of force necessary to make an object start to move from rest, the kinetic friction is the force needed to keep that object moving in that surface, since the car is not starting from rest we are only speaking about kinetic friction.

Kitty [74]3 years ago
6 0

Answer:

<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>

Explanation:

This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc

<em>ALSO</em><em>,</em><em> </em><em>PLEASE DO</em><em> </em><em>MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>

<em>Bonne</em><em> </em><em>journée</em><em> </em><em>;</em><em>)</em><em> </em>

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Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R
vodomira [7]

Answer:

V = IR

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

I\ \alpha\ \frac{V}{R}

Convert the expression to an equation

I\ =\ \frac{V}{R}

Multiply both sides by R to make V the subject

I\ * R\ =\ \frac{V}{R} * R

I\ * R\ =V

Reorder

V = I\ * R

V = IR

<em>Option (a) is correct; Others are not</em>

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector X = 7.8*Cos(50) = 5.01 Km y y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km) so that Tan (\alpha ) = \frac{0.375}{5.01} = 4.28; \alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree) to get how far we use Pythagorean theorem so R^{2} = x^{2}+y^{2} so that R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)

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4 years ago
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you have to substitiute volume of the ballon

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marusya05 [52]

Explanation:

The new volume of water = 25 ml

The old volume of water = 15 ml

The difference = 25 - 15 but what are the units?

Since the question asks for force, the units must start out as 10 mL

In water 1 mL has a mass of 1 gram, so the answer is 10 grams.

Grams are units of mass, not weight. You should convert this into newtons.

10 grams = 1/1000 = 0.01 kg

1 kg has a weight of 9.81 Newtons

0.01 kg has a weight 0.081 Newtons

If you have never seen a Newton before, then the answer is 10 grams

3 0
3 years ago
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