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3 years ago

What would happen happen if a car crashes

Physics

1 answer:

Ierofanga [76]3 years ago

8 0

Answer:

Your body just being crushed from the side. With a side-impact we see much more severe injuries to the thorax and upper-body; you get a lot more rib fractures; a lot more damage to the lungs and internal organs because of the side-impact.

Explanation:

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What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

elena-14-01-66 [18.8K]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

3 0

2 years ago

Read 2 more answers

An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o

Monica [59]

I’m sorry if i took up a lot of space, hope this is a valid approximate answer

6 0

2 years ago

A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?

yanalaym [24]

The first thing you should know for this case is that density is defined as the quotient between mass and volume:
D = M / V
In addition, you should keep in mind the following conversion:
1Kg = 1000g
Substituting the values we have:
D = (23.0 * 1000) / (2920) = 7.88 g / cm ^ 3
answer
the density of the iron plate is 7.88 g / cm ^ 3

8 0

3 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$

y-component of the velocity is: $v_y=4\,*\,2=8\,m/s$

7 0

3 years ago

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c

Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ = 0.23805 N-m

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0

2 years ago

What would happen happen if a car crashes​

What would happen happen if a car crashes