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spayn [35]
3 years ago
10

A 73-kg person in a moving car stops during a car collision in a distance of 0.80 m . The stopping force that the air bag exerts

on the person is 8000 N. Determine the speed of the person before the air bag opens up.
Physics
2 answers:
Vlad [161]3 years ago
6 0

Answer:

13.24m/s

Explanation:

To solve this exercise we go back to the kinematic equations of motion in search of acceleration and the initial speed at which the object was moving.

Once found it will be possible to determine the estimated time.

Our values are:

F=-8000N

m=73Kg

Applying the Second Newton's Law:

a= \frac{F}{m}

a=\frac{8000}{73}

a = 109.58m/s^2

During the collision the distance is 0.8m and there is not final velocity, then

v_f^2 = v^2_0+2ax

0 = v^2_0+2(109.58)(0.8)

v_^2_0 = 175.328

v_0 = 13.24m/s

From the Kinetic Equation we have also

v_f = v_0 - at

0 = 13.24-(109.58)t

t= 0.1208s

Therefore the velocity before the air bag opens up is 13.24m/s in an interval time of 0.12s to open the air bag

fredd [130]3 years ago
3 0

Answer:

 v₀ = 13.24 m / s

Explanation:

Let's use Newton's second law to find the average acceleration during the crash

       F = m a

.       a = F / m

       a = 8000/73

       a = 109.59 m / s²

Now we can use the kinematic equations to find the initial velocity, since when the velocity stops it is zero (v = 0)

       v² = v₀² - 2 a x

       v₀² = 2 a x

       v₀ = √  2 a x

       v₀ = √ (2 109.59 0.80)

       v₀ = 13.24 m / s

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MakcuM [25]

Answer:

F1 = G m1 m2 / R^2        force of attraction

F2 = G m1 m2 / (R/2)^2

F2 / F1 = 4       the force of gravity will be quadrupled

6 0
2 years ago
3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?
Nesterboy [21]

Answer:

I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space

8 0
3 years ago
A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00
Airida [17]

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

7 0
3 years ago
Statistical time division multiplexing does not require the capacity of the circuit to be equal to the sum of the combined circu
aleksklad [387]

Answer:

The answer is True

Explanation:

Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.

In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.

Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.

7 0
3 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
3 years ago
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