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Sedbober [7]
3 years ago
14

Why friction is needed to walk on the ground ?(☆▽☆)(≧▽≦)​

Physics
2 answers:
REY [17]3 years ago
7 0

  • <em>friction acts to grip the ground and prevent sliding.</em>

<h2><em>hope</em><em> </em><em>it</em><em> </em><em>helps!</em></h2>

seraphim [82]3 years ago
4 0

<u>The force of friction is the necessary force which helps us to walk on ground. The friction is generated between the shoes and the road. We press the ground and as a reaction the ground pushes us forward. The force of friction between sole and ground provides the necessary grip which pushes us forward.</u>

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sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he
podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0
3 years ago
7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
trasher [3.6K]

Answer: Car collide with man

Explanation:

Given

Speed of car is u=30\ m/s

Distance of the man from the car is s=55\ m

Reaction time t_r=0.5\ s

Rate of deceleration a_d=-10\ m/s^2

Distance traveled in the reaction time d_o=30\times 0.5=15\ m

Net effective distance to cover d=55-15=40\ m

Distance required to stop the car

\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m

Require distance is more than that of net effective distance. Hence, car collides with the man.

6 0
3 years ago
A trio of students push a 65 kg crate. The first student pushes 31 N [e], the second student pushes 28 N [s] and the third stude
Verdich [7]

Answer:

The free body diagram is attached.

Explanation:

A force of 31[N] to the east, the second force goes to the south and it is equal to 28[N], the third force goes to the west and it is equal to 39 [N].

We can consider the crate as a particle. And all the forces are acting over the particle.

5 0
3 years ago
If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632
ExtremeBDS [4]

Assuming ideal conditions, Boyle's law says that

<em>P₁ V₁ </em>= <em>P₂</em> <em>V₂</em>

where <em>P₁ </em>and <em>V₁</em> are the initial pressure and temperature, respectively, and <em>P₂</em> and <em>V₂</em> are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) <em>V₂</em>

==>   <em>V₂</em> = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

4 0
3 years ago
A 45 Kg object is given a net force of 500 n what is its approximate acceleration
Vanyuwa [196]
Acceleration = force / mass = 500/45 = 11.1 m/s^2
4 0
3 years ago
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