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My name is Ann [436]

3 years ago

6

If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

splacement do you have after this walk?

1 answer:

xeze [42]3 years ago

5 0

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

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A skateboarder with a mass of 60 kg moves with a force of 20 N. What is her acceleration?

Zanzabum

Explanation:

Solution,

Mass(m)= 60 kg
Force (F)= 20 N
Acceleration (a)= ?

We know that,

F=ma
a=F/m
a=20/60
a=0.333 m/s²

So, her acceleration is 0.333 m/s².

4 0

2 years ago

A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth

Gre4nikov [31]

Answer:

The tension is $T = 4326.7 \ N$

Explanation:

From the question we are told that

The total mass is $m = 200 \ kg$

The radius is $r = 5 \ m$

The density of air is $\rho_a = 1.225 \ kg/m^3$

Generally the upward force acting on the balloon is mathematically represented as

$F_N = T + mg$

=> $(\rho_a * V * g ) = T + mg$

=> $T = (\rho_a * V * g ) - mg$

Here V is the volume of the spherical helium filled balloon which is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

=> $V = \frac{4}{3} * 3.142 *(5)^3$

=> $V = 523.67\ m^3$

So

$T = (1.225 * 523.67* 9.8 ) - 200 * 9.8$

$T = 4326.7 \ N$

5 0

3 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago

The atmosphere of Mars is almost all carbon dioxide and the average surface pressure is 610 Pa (as compared with 101,000 Pa on E

Karolina [17]

Answer:

z = 3,737 10⁵ m

Explanation:

a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation

P V = n R T

P = (n r / V) T

We replace

P = (n R / V) T₀ $e^{- C z}$

b) Let's apply this equation in the points

Lower

.z = 0

P₀ = 610 Pa

P₀ = (nR / V) T₀

Higher.

P = 10 Pa

P = (n R / V) T₀ e^{- C z}

We replace

P = P₀ e^{- C z}

e^{- C z} = P / P₀

C z = ln P₀ / P

z = 1 / C ln P₀ / P

Let's calculate

z = 1 / 1.1 10⁻⁵ ln (610/10)

z = 3,737 10⁵ m

4 0

4 years ago

A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player

White raven [17]

Answer:

(a) -472.305 J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by $0.5mv^{2}$

Initial kinetic energy is $0.5\times 69\times 3.7^{2}=472.305 J$

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305 J=-472.305 J

(b)

From fundamental kinematic equation

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

$s=\frac {v^{2}-u^{2}}{-2a}$ but a=\mu g hence

$s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m$

7 0

3 years ago