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Aleonysh [2.5K]
3 years ago
8

A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s

teering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
Physics
1 answer:
egoroff_w [7]3 years ago
4 0

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
Which statement describes a benefit of a model that uses three balls to model the Sun-Earth-Moon system? A. Every aspect of the
Dovator [93]

Answer:

D. It represents a very large, complex system.​

Explanation:

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2 years ago
The major difference between transverse and longitudinal waves is the:
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In transversal wave particles do not oscillate along the line of the wave propagation but oscillate up and down about their mean position as the wave travels.thus transverse wave is the one in which individual particles of the medium move about their mean position in a direction perpendicular to the direction of wave propprgation.
In longitutional waves the individual particles of the medium move in the direction parallel to the direction of the propprgation of the disturbance. The particles do not move from one place to another but the simply oscillate back and forth about their position of the rest.
HOPE IT HELPS.
3 0
3 years ago
Nelle missioni di Apollo sulla Luna,il modulo di comando orbitava a un altitudine di 110 km al di sopra della superficie lunare.
liberstina [14]

Answer:

The period of orbit = 7143.41 s = 119.1 min = 1.984 hours

Il periodo dell'orbita = 7143.41 s = 119.1 min = 1.984 hours

Explanation:

English Translation

In Apollo's missions to the moon, the command module orbited at an altitude of 110 km above the lunar surface. How long did the command module complete an orbit?

Solution

According to Kepler's laws, the square of the period of orbit is proportional to the cube of the radius of orbit.

T² ∝ r³

T² = kr³

where k = constant of proportionality. The constant of proportionality is then given as

k = (4π²/GM)

It's all obtained from equating the gravitational force on the command module by the moon and the circular motion of the command module

Let

G = Gravitational constant

M = mass of the moon

m = mass of the command module

r = radius of orbit

w = angular velocity of the command module

(GMm/r²) = mrw²

(GM/r²) = rw²

(GM/r³) = w²

But angular velocity is given as

w = (2π/T)

w² = (4π²/T²)

(GM/r³) = (4π²/T²)

We then obtain

T² = (4π²/GM)r³

T² = kr³

Mass of moon = M = (7.35 x 10²²) kg Gravitational constant = G = (6.67 x 10⁻¹¹) Nm²/kg²

radius of moon = (1.74 x 10⁶) m

Total radius of orbit = 110 km + (radius of the moon) = 110,000 + (1.74 x 10⁶)

= (1.85 × 10⁶) m

k = (4π²/GM) = (4π² ÷ [(6.67 x 10⁻¹¹) × (7.35 x 10²²)] = (8.059 × 10⁻¹²) kg/Nm²

T² = kr³

T² = (8.059 × 10⁻¹²) × (1.85 × 10⁶)³

T² = 51,028,321.74

T = √(51,028,321.74) = 7,143.41107175

T = 7143.41 s = 119.1 min = 1.984 hours

Hope this Helps!!!

5 0
3 years ago
How many cubic objects of volume 2cm cube can be started in a room of dimension 2m by 3m by 4m​
jenyasd209 [6]

Answer:

12,000,000 boxes

Explanation:

the volume of the room can be found by using the equation for volume of a rectangular box:V=LxWxH

where:

L=2m

W=3m

H=4m

(it doesn't really matter which is which since it is multiplication)

when we multiply our values (2m*3m*4m) we get 24cubic meters

now we need to convert cubic meters to cubic centimeters

each cubic meter is 1,000,000 cubic centimeter we multiply 24 by 1,000,000 and we get: 24,000,000 cubic centimeters (cc)

dividing 24,000,000 by 2 (since each box is 2cc) we get 12,000,000

so, we know we can fit 12,000,000, 2 cubic centimeter boxes in this room

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3 years ago
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