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2 years ago

a cyclist must travel 800 km. how many days will the trip take if the cyclist travels 8/h day at an average speed of 16km/h

2 answers:

7 0

It would b 6 days and 6 hours
You would times 16 by 8 which gives 128km a day. To find how many days you would divide 800 by how many kilometres they cover a day which is 128
So 800/128 =6.25 which converts to 6 days and 6 hours.

Orlov [11]2 years ago

7 0

Answer:at the rate of 16km/h he will travel 16*8=128km per day

800/128=6.25 days

Explanation:

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Which of the following correctly describes the importance of demonstrations to scientific investigation?

sergey [27]

Answer:

A- A demonstration shows how something works, often including models

Explanation:

A demonstration allows, through experimentation, to show how nature works and in that way can include the explanation of scientific theories that explain the set of observed facts, that is, it serves as a demonstration of the underlying scientific principles.

5 0

2 years ago

a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

stiv31 [10]

Answer:

W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

F = W

Where F is the force of the door and W the weight of water

W = mg

We use the concept of density

ρ = m / V

m = ρ V

The volume of the water column is

V = A h

We replace

W = ρ A h g

On the other side the cylinder cover has a pressure

P = F / A

F = P A

We match the two equations

P A = ρ A h g

P = ρ g h

P = 39200 Pa

The weight of the water column is

W = 1000 9.8 4 A

W / A = 39200 kg / m²

3 0

3 years ago

A train is moving in a straight railway where it covered one third of the distance with

Alexxandr [17]

Answer:

<em>The average speed of the train is 45 km/h</em>

Explanation:

<u>Speed</u>

It's defined as the distance (d) per unit of time (t) traveled by an object. The formula is:

$\displaystyle v=\frac{d}{t}$

Let's call x the total distance covered by the train. It covered d1=1/3x with a speed of v1=25 km/h. The time taken is calculated solving for t:

$\displaystyle t_1=\frac{d_1}{v_1}$

$\displaystyle t_1=\frac{1/3x}{25}$

$\displaystyle t_1=\frac{x}{75}$

Now the rest of the distance:

d2 = x - 1/3x = 2/3x

Was covered at v2=75 km/h. Thus the time taken is:

$\displaystyle t_2=\frac{d_2}{v_2}$

$\displaystyle t_2=\frac{2/3x}{75}$

$\displaystyle t_2=\frac{2x}{225}$

The total time is:

$\displaystyle t_t=\frac{x}{75}+\frac{2x}{225}$

$\displaystyle t_t=\frac{3x}{225}+\frac{2x}{225}$

$\displaystyle t_t=\frac{5x}{225}$

Simplifying:

$\displaystyle t_t=\frac{x}{45}$

The average speed is the total distance divided by the total time:

$\displaystyle \bar v=\frac{x}{\frac{x}{45}}$

Simplifying:

$\boxed{\displaystyle \bar v=45\ km/h}$

The average speed of the train is 45 km/h

5 0

2 years ago

a block measuring 20cm by 10cm by 5cm rests on a flat surface. The block has a weight of 3N. Determine the maximum pressure it e

Lady_Fox [76]

Max preassure = force / min area
= 3N / 0.1 x 0.05
= 600N/m(squared)
Copy off of the picture below itll help better, its what someone sent me when i asked this question

4 0

3 years ago

A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force

MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = $\frac{D}{2}$

= $\frac{200}{2}\ m$

= 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

Force F = $\frac{mv^2}{R}$

= $\frac{1500\times (25)^2}{100}\ N$

= 9375 N. [ANS}

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.

6 0

2 years ago