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Leno4ka [110]
3 years ago
12

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 4.60 m/s

in 0.888 s. What are the magnitudes of the linear impulse and the average total force experienced by a 74.0 kg passenger in the car during the time the car accelerates
Physics
1 answer:
serg [7]3 years ago
3 0

Answer:

(a).The value of force applied on the car during acceleration F = 383.32 N

(b).The value of impulse on the car  I = 340.4 N-sec

Explanation:

Acceleration of the car  a = \frac{4.6}{0.888} = 5.18 \frac{m}{s^{2} }

Mass of the passengers = 74 kg

Total force experienced by the car = mass  × acceleration

⇒ F = m × a

Put the values of mass & acceleration in the above formula we get

Force F = 74 × 5.18

⇒ F = 383.32 N

This is the value of force applied on the car during acceleration.

linear Impulse is given by I = F × Δt

Where F = force applied on the car

Δt  = time interval upon which the force is applied on the car

⇒ I = 383.32 × 0.888

⇒ I = 340.4 N-sec

This is the value of impulse on the car.

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Answer:

The acceleration is equal to the net force divided by the mass. If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

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A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
3 years ago
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max2010maxim [7]

Answer:

2.5 m/s

Explanation:

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E = PE + KE

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v = 2.5 m/s

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