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Advocard [28]
1 year ago
14

This is the third time I’m asking, please, On a wet road, is a higher coefficient of friction on the tires safer or a lower one

and pls explain I don’t get it
Physics
1 answer:
jenyasd209 [6]1 year ago
4 0

Answer:

higher is safer

Explanation:

because it is a wet slippery surface, you would need more friction on the tires, to get more traction in the slippery wet road, if you had low friction you would not move anywhere and or could swerve off somewhere

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How are frequency and were peroid related?
hodyreva [135]

Answer:

Frequency and period are distinctly different, yet related, quantities. Frequency refers to how often something happens. Period refers to the time it takes something to happen. Frequency is a rate quantity.

7 0
3 years ago
Question 7
Drupady [299]
Power = energy/time=20/4=5.0
6 0
3 years ago
Read 2 more answers
An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e
Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0
3 years ago
You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m
slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock m=500 gm

diameter of circle d=2.2 m

radius r=\frac{2.2}{2}=1.1 m

At highest Point

mg+N=\frac{mv^2}{r}

At highest Point N=0 because mass is just balanced by centripetal Force

thus mg=\frac{mv^2}{r}

v=\sqrt{gr}

v=\sqrt{9.8\times 1.1}

v=\sqrt{10.78}

v=3.28 m/s

6 0
3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
3 years ago
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