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3 years ago

During a car crash, why are airbags useful? 0.They decrease the amount of impulse applied to the you.

Physics

2 answers:

Vera_Pavlovna [14]3 years ago

3 0

Answer:

They increase the time it takes to slow you down, which decreases the force applied to you.

Explanation:

Hope it helps :)))

Bumek [7]3 years ago

3 0

Answer:

They increase the time it takes to slow you down, which decreases the force applied to you.

Explanation:

Air bags can only deploy at around 8-14mph

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An "energy bar" contains 26 g of carbohydrates. For the steps and strategies involved in solving a similar problem, you may view

masha68 [24]

To solve this problem we will apply the definition of Power and Speed. In turn, we will consider that one gram of carbohydrate, according to numerous scientific studies, contributes around 17kJ of energy. Therefore, if this were true, the total energy of 26 grams would be

$E = (26)(17000) = 4.42*10^5J$

Power can be described as the amount of energy applied at a given time, that is,

$P = \frac{E}{t} \rightarrow t = \frac{E}{P}$

$t = \frac{4.42*10^5}{380}$

$t = 1.16*10^3s$

The speed is described as the distance traveled in a certain time, and its units in international system is m / s, converting and replacing we will have

$v = 5km/h(\frac{1000m}{1km})(\frac{1h}{3600s})$

$v = 1.388m/s$

Now,

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

The distance is,

$d = vt$

$d = (1.388)(1.16*10^3)$

$d = 1610.08m$

Therefore the distance walked is 1610.08m

7 0

3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

PLEASE HELP ME!

Damm [24]

<span>In physical cosmology, cosmic inflation, cosmological inflation, or just inflation, is a theory of exponential expansion of space in the early universe. The inflationary epoch lasted from 10−36 seconds after the conjectured Big Bang singularity to sometime between 10−33 and 10−32 seconds after the singularity.
I hope this helps!!!
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4 0

3 years ago

Read 2 more answers

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

$U = \dfrac{1}{2}kx^2$