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klemol [59]
2 years ago
9

On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the

circuit so that, maintaining the same speed, the intervals between them will decrease by 1 5
Physics
1 answer:
Sphinxa [80]2 years ago
4 0

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

d=\frac{L}{12} (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

d'=(1-\frac{1}{5})d=\frac{4}{5}d

And the number of trams will become

12+n

So eq.(1) will become

\frac{4}{5}d=\frac{L}{n+12} (2)

And substituting eq.(1) into eq.(2), we find:

\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3

Learn more about distance and speed:

brainly.com/question/8893949

#LearnwithBrainly

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The average kinetic energy of the molecules of an ideal gas at 10∘C has the value K10. At what temperature T1 (in degrees Celsiu
Westkost [7]

Answer:

A) T1 = 566 k = 293°C

B) T2 = 1132 k = 859°C

Explanation:

A)

The average kinetic energy of the molecules of an ideal gas is givwn by the formula:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy

K = Boltzman Constant

T = Absolute Temperature

At 10°C:

K.E = K10

T = 10°C + 273 = 283 K

Therefore,

K10 = (3/2)(K)(283)

FOR TWICE VALUE OF K10:

T = T1

Therefore,

2 K10 = (3/2)(K)(T1)

using the value of K10:

2(3/2)(K)(283) = (3/2)(K)(T1)

<u>T1 = 566 k = 293°C</u>

<u></u>

B)

The average kinetic energy of the molecules of an ideal gas is given by the formula:

K.E = (3/2)KT

but K.E is also given by:

K.E = (1/2)(m)(vrms)²

Therefore,

(3/2)KT = (1/2)(m)(vrms)²

vrms = √(3KT/m)

where,

vrms = Root Mean Square Velocity of Molecule

K = Boltzman Constant

T = Absolute Temperature

m = mass

At

T = 10°C + 273 = 283 K

vrms = √[3K(283)/m]

FOR TWICE VALUE OF vrms:

T = T2

Therefore,

2 vrms = √(3KT2/m)

using the value of vrms:

2√[3K(283)/m] = √(3KT2/m)

2√283 = √T2

Squaring on both sides:

(4)(283) = T2

<u>T2 = 1132 k = 859°C</u>

8 0
2 years ago
A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k
9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg  : mass of the  skier

θ =30°  :angle θ of the ski slope  with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W=  92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N=  0.10*780.8 N  

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 =  (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

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2 years ago
You are playing a game called "Will It Float?" In this game, you are given a large, square can of tuna. If you know the density
Delicious77 [7]
The only information you would need to decide if the can will float is the density of the can, which requires knowing the mass and volume. If the density of the can is less than one, the can will float. if it is greater than one, it will not float, as water's density is one.
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2 years ago
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An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

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2 years ago
A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

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7 0
2 years ago
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