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3 years ago

what person is responsible for providing short term care for people being transported to the hospital

Physics

2 answers:

devlian [24]3 years ago

7 0

Answer: First Aider

Explanation:

A person responsible for providing short term care for people being transported to the hospital is a first aider. This people provide first aid to victims from the point of incident till the hospital where such victims will be properly taken care of. They only provide first treatment to the victim prior to being taken to the hospital.

Firstly Aid is therefore defined as the first treatment administered to a patient before being taken to the hospital.

andre [41]3 years ago

6 0

The ambulance people

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A car starts from the state of xestIf its velocity becomes 70 km/hr in 6 minutes, i) what is the accordine acceleration of F the

Alexxandr [17]

Answer: $3.5\ km$

Explanation:

Given

Car starts from the state of rest and acquires a velocity of $70\ km/hr$ in 6 minutes

Final velocity in m/s is $v=70\approx 19.44\ m/s$

Using equation of motion

$v=u+at\\\Rightarrow 19.44=0+a(6\times 60)\\\Rightarrow a=0.054\ m/s^2$

Distance covered in 360 s

$\Rightarrow v^2-u^2=2as\\\Rightarrow 19.44^2-0=2\times 0.054\times s\\\Rightarrow s=3500.64\ m\approx 3.5\ km$

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3 years ago

What ocean depth would the volume of an aluminium sphere be reduced by 0.10%

yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

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3 years ago

Hi gir_ls join nkd-mbja-nuj

GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

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3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0

3 years ago