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lora16 [44]
3 years ago
11

A dog walks a distance of 55.5 meters in 120 seconds. What was its speed?

Physics
1 answer:
Nitella [24]3 years ago
8 0

Answer: ...

Explanation: 1: The dog is walking 0.4625MPS (Meters per second. M/S)

To calculate speed you divide the distance by time so 55.5 Meters/120 Seconds is 0.4625 MPS

2: The car traveled 15 KM

To calculate speed you need to multiply the speed by time. (Since it's multiplication it can be put in any order) So 600 Secondsx25 Meters is 15,000 M which is 15 KM.

3: It would take 8 Minutes and 20 Seconds to travel 1500 Meters.

To calculate the time you multiply Speed by distance (Order doesn't count.)

So 1500 Metersx3 MPS is 8 Minutes and 20 seconds.

4: It was going 2.7775 MPS.

To calculate the speed you divide the distance over time. So 9999 Meters/3600 seconds is 2.7775 MPS.

5: It was going 1.1222... (Infinite)

To calculate the speed you divide the distance over time

So 8080 Meters/ 7200 Seconds is 1.222...

6: It would cover 1.375 Meters.

To calculate the Distance you multiply Speed by time (Order doesn't count.)

So 10.25 Secondsx5.5 MPS is 1.375 Meters.

7: It would take 1 Minute and 31 Seconds to travel 10 Meters.

To calculate the time you multiply Speed by distance (Order doesn't count.)

So 10 Metersx0.11 MPS is 1 Minute and 31 seconds

Hopes this helps because this took over 10 minutes.

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7 0
3 years ago
Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

7 0
3 years ago
ANSWER ASAP PLZZZZ!!!!!!!!
igor_vitrenko [27]

Answer: your correct answer is a i took the test

Please i need brainlist i need one more and i level up :)

6 0
3 years ago
Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds 5.0
Tems11 [23]
A. Average speed is weighted mean (1 × 2 + 2 × 3 + 3 × 5 + 4 × 7 + 3 × 9 + 2 × 12.5)/15 = (2 + 6 + 15 + 28 + 27 + 25)/15 = 103/15 = 6.867 b. RMS is square root of 1/15 times sum of squares of speeds Sum of squares is 4 + 9 + 9 + 25 + 25 + 25 + 49 + 49 + 49 + 49 + 81 + 81 + 81 +156.25 + 156.25 = 848.5 
c. RMS speed = √(848.5/15) = 7.521 
Most likely the speed is the peak in the speed distribution, which is 7.
5 0
3 years ago
A 55.4 g sample of water at 99.61 °C is placed in a constant pressure calorimeter. Then, 23.4 g of zinc metal at 21.6 °C is adde
Zolol [24]

Answer:

The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C

Explanation:

From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture

Heat loss by water = Heat gain by zinc metal

m1C1(T1 - T3) = m2C2(T3 - T2)

m1 is mass of water = 55.4 g

C1 is specific heat capacity of water = 4.2 J/g.°C

m2 is mass of zinc metal = 23.4 g

C2 is specific heat capacity of zinc metal

T1 is the initial temperature of water = 99.61 °C

T2 is the initial temperature of zinc metal = 21.6 °C

T3 is the final temperature of the mixture = 96.4 °C

55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)

746.9028 = 1750.32C2

C2 = 746.9028/1750.32 = 0.427 J/g.°C

3 0
3 years ago
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