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aksik [14]
3 years ago
8

If a Mack truck and Honda Civic have a head-on collision, upon which vehicle is the impact force greater?

Physics
1 answer:
goblinko [34]3 years ago
8 0

Answer:

Honda Civic

Explanation:

As per the structural design and purpose of use for vehicle, Both are different from one another in various terms such as size, strength and design.

Mack trucks are very huge trucks especially designed for loading heavy objects. They are even used to carry heavy consignments such as cars.

Whereas Honda civic has a sleek structural design and have a very fragile body as it is sedan car, so if there will be a head-on collision between these two vehicles, certainly there will be a huge impact force on civic and it will even smash into pieces if any of the vehicle have high speed.

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Explanation:

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If you wanted to change liquid water into solid ice you would _____ it
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you would freeze it!

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How strong an electric field is needed to accelerate electrons in an X-ray tube from rest to one-tenth the speed of light in a d
Bond [772]

Answer:

E= 50.1*10³ N/C

Explanation:

Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:

vf^{2} -vo^{2}  = 2*a*x

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a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2

According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.

Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:

F = q*E (1)

In this case, q=e= 1.6*10⁻19 C

But this force, can be expressed in this way, according Newton's 2nd Law:

F = m*a (2) ,

where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.

From (1) and (2), we can solve for E, as  follows:

E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C

⇒ E = 50.1*10³ N/C

3 0
4 years ago
A projectile is launched with an initial velocity 60m/s at an angle 60° to the vertical. What the magnitude of it's displacement
emmasim [6.3K]

Answer:

the magnitude of the displacement after 5s is 137.31 m.

Explanation:

Given;

initial velocity of the projectile, u = 60 m/s

angle of projection, θ = 60°

time of motion, t = 5s

the vertical component of the velocity, u_y= u\ sin \theta = 60sin(60^0)

The magnitude of the displacement after 5s is calculated as;

h = u_yt -\frac{1}{2} gt^2\\\\h = 60sin (60^0)\times 5 - \frac{1}{2} (9.8)(5)^2\\\\h = 259.81-122.5\\\\h = 137.31 \ m

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