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3 years ago

The mass of the bicycle and rider is 60 kg and the total area of the tyres in contact with

Physics

1 answer:

AleksAgata [21]3 years ago

3 0

Answer:

a little

Explanation:

First of all, it's not how you spell "tyres", it is tires.

Second of all, you already know the Mass so what you need to find out now is contact the road. You Know that your number and letter are squared so that would turn into 6m x 2.4. Then you do the math do continue on to finish it. Have a great day!! Good luck with the answer!!

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Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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#LearnwithBrainly

7 0

4 years ago

If 40.0 kj of energy are absorbed by .500 kg of water 10 degrees c, what is the final temperature of the water?

anastassius [24]

To solve the problem, we must use the following equation:

$Q=mC_s (T_f -T_i)$

where

Q is the amount of heat energy absorbed by the water

m is the mass of the water

Ti and Tf are the initial and final temperature

Cs is the specific heat capacity of the water

The data we have in this problem are:

Q=40.0 kJ

$C_s =4.186 kJ/kg^{\circ}C$

$T_i=10^{\circ}C$

m=0.500 kg

Substituting the data into the equation and re-arranging it, we find

$T_f = T_i + \frac{Q}{mC_s}=10^{\circ}+\frac{40.0 kJ}{(0.500 kg)(4.186 kJ/kg^{\circ}}=29.1^{\circ}C$

So the final temperature of the water will be 29.1 degrees.

8 0

3 years ago

A ship's boiler steam comes out at 112 C and pushes through the system, exiting into the condenser,

solniwko [45]

Answer:

D: 0.239

Explanation:

Equation for ideal efficiency is;

η = 1 - (T_c/T_h)

We are told that;

steam comes out at 112° C. Thus, T_h = 112°C. Converting to Kelvin gives; T_h = 112 + 273 = 385 K

The one exiting into the condenser is kept at 20°C. Thus; T_c = 20 + 273 = 293 K

Thus;

η = 1 - (293/385)

η = 0.239

5 0

3 years ago

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. if the passengers are to ex

sleet_krkn [62]

The answer is "156.6 m/s".

This is how we calculate this;

-N + mg = ma = mv²/r

For "weightlessness" N = 0, so

0 = mg - mv²/r

g - v²/r = 0

v =√( gr)
g = 9.8 and r = 2.5km = 2500 m

v = √(9.8 x 2500)

= 156.6 m/s

8 0

3 years ago

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, $a=1.8\ m/s^2$