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irina [24]
3 years ago
12

⦁ A car going 50 m/s is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large an average force is exer

ted by seatbelts on a 90 kg passenger as the car is stopped?
Physics
1 answer:
gtnhenbr [62]3 years ago
8 0

Answer:

the average force exerted by seatbelts on the passenger is 5625 N.

Explanation:

Given;

initial velocity of the car, u = 50 m/s

distance traveled by the car, s = 20 m

final velocity of the after coming to rest, v = 0

mass of the passenger, m = 90 kg

Determine the acceleration of the car as it hit the pile of dirt;

v² = u² + 2as

0 = 50² + (2 x 20)a

0 = 2500 + 40a

40a = -2500

a = -2500/40

a = -62.5 m/s²

The deceleration of the car is 62.5 m/s²

The force exerted on the passenger by the backward action of the car is calculated as follows;

F = ma

F = 90 x 62.5

F = 5625 N

Therefore, the average force exerted by seatbelts on the passenger is 5625 N.

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We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth
Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = \frac{m}{m+M} \ v_o

for when n balls have collided

          v = \frac{m}{n \ m + M}  v₀

Now let's analyze the case of the bouncing ball (elastic)

     

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = \frac{m-M}{m + M}   v₀

we substitute in equation 1 to find v₂

            \frac{M}{m}  v₂ = v₀ -  \frac{m-M}{m+M}   v₀

            v₂ = \frac{m}{M}  ( 1 - \frac{m-M}{m+M} ) \ v_o

            v₂ = \frac{m}{M}  ( \frac{2M}{m+M} ) \ \ v_o

            v₂ = \frac{2m}{m +M}  \ v_o  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0
2 years ago
Assuming there are no accidents or delays, the distance that a car travels down the interstate
Ganezh [65]

Explanation:

The distance that a car travels down the interstate  can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

or

D=6.75\times 10^5\ m

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

or

D=1.12\times 10^6\ m

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

or

D=1.68\times 10^6\ m

Hence, this is the required solution.

7 0
3 years ago
10 minutes on the stove, the water molecules in the pot are in two different
DaniilM [7]

Answer:

20 molcues

Explanation:

8 0
3 years ago
According to the equation for the Balmer line spectrum of hydrogen, a value of n = 3 gives a red spectral line at 656.3 nm, a va
nirvana33 [79]

Answer:

E3 = 3.03 10⁻¹⁶ kJ,  E4 = 4.09 10⁻¹⁶ kJ and  E5 = 4.58 10⁻¹⁶ kJ

Explanation:

They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units

     n = 3        λ = 656.3 nm = 656.3 10⁻⁹ m

     n = 4        λ = 486.1 nm = 486.1 10⁻⁹ m

     n = 5        λ=434.0 nm = 434.0 10⁻⁹ m

Let's use the Planck equation

     E = h f

The speed of light equation

   c = λ f

replace

    E = h c /λ

Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s

Let's calculate the energies

     E = 6.63 10⁻³⁴ 3 10⁸ / λ

     E = 19.89 10⁻²⁶ /λ

n = 3

    E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹

    E3 = 3.03 10⁻¹⁹ J

    1 kJ = 10³ J

    E3 = 3.03 10⁻¹⁶ kJ

n = 4

    E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹

    E4 = 4.09 10⁻¹⁹ J

    E4 = 4.09 10⁻¹⁶ kJ

n = 5

    E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹

    E5 = 4.58 10⁻¹⁹ J

    E5 = 4.58 10⁻¹⁶ kJ

7 0
3 years ago
In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of t
a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

        = 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵  = 1000 x 9.81 x h

h = 65.85 meters

7 0
3 years ago
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