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Wittaler [7]
4 years ago
6

In a bike race drake drove his bike over the 1500 meter course from start to finish in 50 seconds what was his average speed

Physics
2 answers:
Yuri [45]4 years ago
5 0

Answer:

The answer is Average speed is equal to 30m/sec

Step by Step Explanation:

1- The average speed is the relationship between the displacement that a body made and the total time it took to perform it.

2- The formula that allows to calculate the average velocity is V = ΔX/ΔT where (V) is the average speed, (ΔX) is the displacement and (ΔT) is the time interval.

3- Knowing this beforehand, we can proceed to solve the problem by replacing the given values.

v = \frac{1500m}{50sec} \\v = 30\frac{m}{sec}

Maslowich4 years ago
4 0
30m/s. ................
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Answer:

Determine how rare a diamond is

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3 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch ve
Oksana_A [137]

Answer:

gs = 0.6 m/s^2

Explanation:

Given data:

velocity = 12 m/s

height s = 12t -(1/2) g_s t^2

Given velocity is the derivatives of  height

v(t) = \frac{d}{dt} s(t)

      = \frac{d}{dt}(12t -\frac{1}{2} g_s t^2)

      = 12 - g_s t

when velocity tend to 0 , maximum height is reached

v(t) = 12 - g_s t

0 = 12 - g_s t

g_s = \frac{12}{t}

at t = 20 sec ball reached the max height, so

g_s = \frac{12}{20} = 0.6 m/s^2

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How do I solve using the formula:<br><br> Vf^2=vo^2+2gh<br><br> ^2 means square
PtichkaEL [24]

your second line ... minus 2gh ...

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3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

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where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

C=372.27J/^oC=3.7\times 10^2J/^oC

Therefore, the heat capacity of the calorimeter is, 3.7\times 10^2J/^oC

7 0
3 years ago
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