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4 years ago

A passenger jet flies from one airport to another 1293 miles away in 2.1hours . Find average speed

Physics

1 answer:

JulsSmile [24]4 years ago

6 0

The average speed is 615.7 mph (275.2 m/s)

Explanation:

The average speed of an object gives a measure of how fast an object is moving over a certain time interval. It is a scalar quantity, and it is calculated as follows

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

where

v is the average speed

d is the distance covered

t is the time interval

For the jet in this problem, we have:

d = 1293 mi is the distance covered

t = 2.1 h is the time interval

Therefore, the average speed is:

$v=\frac{1293}{2.1}=615.7 mph$

We can also convert into SI units (m/s), keeping in mind that:

$1 mi = 1609 m\\1 h = 3600 s$

And so

$v=615.7 \frac{mi}{h}\cdot \frac{1609 m/mi}{3600 s/h}=275.2 m/s$

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Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?

OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0

3 years ago

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?

lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

F₁ = 225.4 N

F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

∑ F = 0

∑ τ = 0

Where F are the forces and τ the torques.

The torque is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

F₂ 2 - W₁ 1 - W₂ 1.5 = 0 $\frac{W_1 \ 1 + W_2 \ 1.5}{2}$

Let's calculate F₂

F₂ = $\frac{W_1 \ 1 + W_2 \ 1.5 }{2}$

F₂ = (m g + M g 1.5)/ 2

F₂ = $\frac{(12 + 68 \ 1.5 ) \ 9.8}{2}$

F₂ = 558.6 N

We substitute in the translational equilibrium equation.

F₁ = W₁ + W₂ - F₂

F₁ = (m + M) g - F₂

F₁ = (12 +68) 9.8 - 558.6

F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

F₁ = 225.4 N

F2 = 558.6 N

Learn more here: brainly.com/question/12830892

5 0

2 years ago

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

Can’t someone help me

Aleonysh [2.5K]

Answer:

Explanation:

Objects with the same charge (Ex. North and north) repel each other.

3 0

3 years ago