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vaieri [72.5K]
3 years ago
8

You are working during your summer break as an amusement park ride operator. The ride you are controlling consists of a large ve

rtical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (Fig. P6.7). The coefficient of static friction between a person of mass m and the wall is ms, and the radius of the cylinder is R. You are rotating the ride with an angular speed v suggested by your supervisor. (a) Suppose a very heavy person enters the ride. Do you need to increase the angular speed so that this person will not slide down the wall? (b) Suppose someone enters the ride wearing a very slippery satin workout outfit. In this case, do you need to increase the angular speed so that this person will not slide down the wall?
Physics
1 answer:
sveta [45]3 years ago
8 0

Answer:

A

Explanation:

because if a very heavy person ride they can slow down the ride

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What is the speed of a commercial jet which travels form New York to Los Angeles (4800) in 6 hours
Colt1911 [192]

Answer:

Speed = 800km}/hr

Explanation:

Given

Distance = 4800km

Time = 6hr

Required

Determine the speed of the jet

The speed is calculated as:

Speed = \frac{Distance}{Time}

Substitute 4800 km for Distance and 6hr for Time

Speed = \frac{4800km}{6hr}

Speed = 800km}/hr

<em>Hence, the speed of the commercial jet is 800km/hr</em>

8 0
2 years ago
Which best describes the law of conservation of mass? The coefficients in front of the chemicals in the reactants should be base
likoan [24]

Answer:

Explanation:

1. when both sides of the reactants and the products are equal

like h+oh ⇒ h20

hydrogen has 2 atoms on both sides and oxygen  has one  on both sides

2. no they are put to balance the equation

3. nope they are treated equally as all the other states

4. no if u are talking about a formula unit like NaCl for example it is aqueous not each element taken on its own if u are talking about just elements then i said before all states are treated equally

5. yup

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2 years ago
How far did a car travel if it was on the road for 55.2s and it traveled at an average speed of 55m/s
pantera1 [17]
The car traveled 1.00363 kilometers in the 55.2s and the speed of 55m/s
3 0
2 years ago
A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x
makkiz [27]

Answer:

A)Magnitude of the net electric field at the origin due q₁ and q₂

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

β=3.91°

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

Data

q₁ = -4 nC = -4*10⁻⁹ C

q₂ = +6nC =+ 6*10⁻⁹ C

k = 9*10⁹ N*m²/C²

d_{1} =\sqrt{0.6^{2}+0.8^{2}  } =1 m

d₂ = 0.6 m

Graphic attached

The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:

E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.

E₂: Total field at point  x=0 , y=0  due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.

Calculation of the electric field at the origin of x-y coordinates due to the charge q₁

E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C

Components (x-y) of the field due to q1:

E₁x=E₁*cosα = 36*(0.6/1)  = 36*(0.6) = 21.6 N/C

E₁y=E₁*sinα  =  36*(0.8/1) = 36*(0.8) =28.8 N/C

Calculation of the electric field at the origin of x-y coordinates due to the charge q₂

E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C

Calculation of the electric field components at the origin of x-y coordinates

E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)

E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C

E₀y= E₁y = 28.8 N/C

A )Magnitude of the net electric field at the origin due q₁ and q₂

E_{o} =\sqrt{128.4^{2}+28.8^{2}  }

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

\beta =tanx{-1} \frac{E_{oy} }{Eox}

\beta =tan^{-1} \frac{28.8}{128.4}

β=3.91°

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In an atom is the number of protons greater, less than or equal to the number of electrons?
Igoryamba

Answer:

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Explanation:

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