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Alchen [17]
2 years ago
11

A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c

oasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.
Physics
1 answer:
Serjik [45]2 years ago
3 0

Solution:

At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s  

Initial velocity = 75 * 5/18 = 20.8 m/s  

The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.  

Now according to the equation determine the distance.  

vf^2 = vi^2 + 2 * a * d  

a = -0.5 m/s^2  

0 = 20.8^2 + 2 * -0.5 * d  

so d = 431.64 m  

since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.  

vf = vi + a * t  

0 = 20.8 – 0.5 * t  

Solve for t = 41 seconds  

(c) the distance travels by it during the first and fifth second are.  

d = vi * t + ½ * a * t^2  

d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m  

The easiest way to the distance for the 5th second is:

d = vi * t + ½ * a * t^2, a = -0.5  

d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m  

d6 = 20.8 * 6 – ½ * 0.5 * 6^2 =  106.8m

d6 – d5 = 15.3 m  

this is the required solution.


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Answer:

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

Explanation:

Given data

Φ = 5.32 eV

to find out

the longest wavelength

solution

we know that

hf = k(maximum) +Ф   ...............1

here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

so hf = 0

and hc/ λ = +Ф

so λ = hc/Ф  ................2

now put value hc = 1240 ev nm and Φ = 5.32 eV

so hc = 1240 / 5.32

hc = 233 nm

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

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3 years ago
The state of matter is an example of a physical property. Why is the state of matter considered aphysical property, even though
lora16 [44]
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3 years ago
En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia
sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

  • Velocidad = 50 m/s
  • Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

Sustituyendo en la fórmula, tenemos;

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7 0
3 years ago
A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

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