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2 years ago

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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

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:

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Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0

east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero

distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m

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2 years ago

You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.

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Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

Mass = 60 Kg

Energy = 5000 J

We need to calculate the specific heat

Using formula of energy

$Q= mc\Delta T$

$c =\dfrac{Q}{m\Delta T}$

Put the value into the formula

$c=\dfrac{5000}{60\times(37-13)}$

$c=3.47222\ J/kg^{\circ}C$

Hence, The specific heat is 3.47222 J/kg°C.

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2 years ago

Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b

zhuklara [117]

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence = 38.0°

For yellow light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}$

${sin\theta_2}=0.9377$

Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}$

${sin\theta_2}=0.9395$

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

${sin\theta_{critical}}=\frac {n_r}{n_i}$

Where,

${\theta_{critical}}$ is the critical angle

$n_r$ is the refractive index of the refractive medium.

$n_i$ is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

${sin\theta_{critical}}=\frac {1}{1.523}$

The critical angle is = sin⁻¹ 0.6566 = 41.04°

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2 years ago

8. Rubbing a plastic bag and a balloon with a cloth gives both objects a net negative charge. The balloon's

Dafna1 [17]

Answer:

0.214 m

Explanation:

In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.

Therefore, we can write:

$k\frac{q_1 q_2}{r^2}=mg$

where

k is the Coulomb constant

$q_1=-1\cdot 10^{-10}C$ is the charge on the balloon

$q_2=-1\cdot 10^{-5} C$ is the charge on the bag

r is the separation betwen the bag and the balloon

$m=0.02 g=2\cdot 10^{-5} kg$ is the mass of the bag

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for r, we find the distance at which the bag must be held:

$r=\sqrt{\frac{kq_1 q_2}{mg}}=\sqrt{\frac{(9\cdot 10^9)(-1\cdot 10^{-10})(-1\cdot 10^{-5})}{(2\cdot 10^{-5})(9.8)}}=0.214 m$

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2 years ago

Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

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2 years ago