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2 years ago

Pls help guys this is my 7th time posting XD

Physics

2 answers:

borishaifa [10]2 years ago

5 0

Answer:

Explanation:

initial velocity u = 54 km/h = 15 m/s

final velocity v = 18 km/h = 5 m/s

distance s = 10 m

1. v^2 = u^2 + 2as

5^2 = 15^2 + 2a × 10

25 = 225 + 20a

25 - 225 = 20a

20 a = -200

a = -200/2

a = -100m/s^2

∴Deacceleration = -100m/s^2

2. v = u + at

5 = 15 -100t

5-15 = -100t

-10 = -100t

t = 100 / 10

∴t = 10 sec

Total distance covered by the car 10 m

zysi [14]2 years ago

5 0

To me, the hardest part of this whole thing is keeping the units straight. We're starting out with information given to us in kilometers, hours, and meters, and we have to come up with answers in m/s² , seconds, and meters.

When I worked this problem, I jumped right in without thinking, and I immediately got bogged down when I had to go off to the side and convert some units.

Now I know better. THIS time, before we get all tangled up trying to solve anything, let's get clever and change everything to m/s right now !

(54 km/hour) · (1,000m/km) · (1 hour/3600 sec) = 15 meters/second

(18 km/hour) · (1,000 m/km) · (1 hour/3600 sec) = 5 meters/second

NOW I'll betcha it's gonna be about 70% faster and easier !

i). Acceleration = (change in speed / time for the change)

We know the distance, but not the time. I know there's a formula for it, but I've learned so many formulas during my lifetime that I can't remember ALL of them. So I just memorize some of them, and I work things out from the formulas that I know. Here's how I do time:)

Average speed during the given slow-down = (1/2)·(15+5) = 10 m/s

Distance covered during the given slow-down = 10 m.

Time = (distance) / (average speed) = (10m) / (10 m/s) = 1 second

Acceleration = (change in speed) / (time for the change)

Acceleration = (5 m/s - 15 m/s) / (1 second)

Acceleration = (-10 m/s) / (1 second)

Acceleration = -10 m/s² (or 'Deceleration' = 10 m/s² )

_____________________________________________

For parts ii). and iii)., there's a big shift in the question.

It only gave you the slow-down from 54 to 18 km/hr for the purpose of calculating the deceleration. NOW, for the rest of the answers, it's talking about a complete stop ... 0 m/s .

____________________________________________

ii). Time required to stop = (initial speed) / (deceleration)

Time to stop from 54 km/hr = (15 m/s) / (10 m/s²)

Time to stop = 1.5 seconds

iii). Distance covered = (average speed) · (time to stop)

Distance covered = (7.5 m/s) · (1.5 sec)

Distance covered = 11.25 meters

OR ... use the official formula:

Distance = (1/2) · (acceleration) · (time² )

Distance = (1/2) · (10 m/s²) · (1.5 sec)²

Distance = 11.25 meters yay !

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Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

8 0

2 years ago

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

3 0

2 years ago

El coeficiente de variación de la resistencia con la temperatura del carbón es -0.0005/°c.Si la resistencia de una resistencia d

Doss [256]

Answer:

R (120) = 940Ω

Explanation:

The variation in resistance with temperature is linear in metals

ΔR (T) = R₀ α ΔT

where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC

let's calculate

ΔR = 1000 (-0,0005) (120-0)

ΔR = -60 Ω

ΔR = R (120) + R (0) = -60

R (120) = -60 + R (0)

R (120) = -60 + 1000

R (120) = 940Ω

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never [62]

Answer:

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tia_tia [17]

Answer: Cell's nucleus

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DNA is coiled into chromosomes in a cell's nucleus. It has a shape like a double helix. It is twisted in the form of spirals.

Hence, DNA is coiled into chromosomes in a cell's Nucleus.

6 0

3 years ago

Read 2 more answers

Pls help guys this is my 7th time posting XD​

Pls help guys this is my 7th time posting XD