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Alex Ar [27]
3 years ago
12

Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri

es.
Will give brainliest + 5 stars!
Physics
1 answer:
ch4aika [34]3 years ago
8 0
The larger the mass the mass and the speed, the larger the force it would produce if it crashes something. Some car crashes produce minor injuries because maybe this car is small and runs at very slow speed which would mean less force when in impact with another. However, if a big truck crashes, it is expected to produce a larger force causing catastrophic injuries.
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A 50.0 kg child stands at the rim of a merry-go-round of radius 1.95 m, rotating with an angular speed of 2.80 rad/s.
eimsori [14]

Answer:

a) ac= 15.3 m/s² b) 765 N c) 1.56

Explanation:

a) For the child standing in the rim of the merry-go-round, there exists a force, called centripetal force, that keeps her moving in a circular path, and not going in a straight line at constant speed, as Newton's first law dictates.

The acceleration cosequence of this force is called centripetal acceleration, and it can be showed be equal to the following expression:

ac = ω²*r, where ω is the angular speed, and r is the radius.

Replacing by the givens:

ac= (2.8)² (rad/sec)² * 1.95 m = 15.3  m/s²

b) As we were talking about the centripetal force, we realize that this can't be some type of  an unknown force, so, it must  be one than can be identifiable.

Looking at the external forces on the child, the only one that is applied in horizontal plane, is the friction force.

So, this friction force, is just the centripetal force we were talking about.

The mimimum force between her feet and the floor of the carousel, is just the centripetal force:

Fc = m*ac = 50.0 kg * 15.3 m/s² = 765 N

c) The friction force can take any value, in order to be equal to the centripetal force,  up to a limit value, beyond which , the child will move. This limit value can be expressed as follows:

Fs max = μs*N  = μs*m*g

⇒ m*ac = μs*m*g

Simplifying  common terms, we can solve for the coefficient of static friction ,  μs, as follows:

μs = 15. 3 m/s² / 9.8 m/s² = 1.56

8 0
3 years ago
A medicine ball has a mass of 5 kg and is thrown with a speed of 2 m/s. what is its kinetic energy?
Fantom [35]
The kinetic energy of the ball is given by:
K= \frac{1}{2}mv^2
where m=5 kg is the mass of the ball and v=2 m/s is its speed. Substituting these numbers, we find the kinetic energy:
K= \frac{1}{2} (5kg)(2m/s)^2=10 J
7 0
3 years ago
A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav
Agata [3.3K]

Answer:

The tension in the string is T = 1.49*10^{-6}N.

Explanation:

For a string with tension T and linear density \mu_d carrying a transverse wave at speed v it is true that

v = \sqrt{\dfrac{T}{\mu_d} }

solving for T we get:

T = \dfrac{v^2}{\mu_d}.

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s

And it's linear density (mass per unit length) is

\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m

Therefore, the tension in the cord is

T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.

\boxed{T = 1.5*10^{-6}N}

or in micro newtons

T =1.5\mu N

4 0
3 years ago
Have thick walls<br>Chamaer o the heart<br>which​
yulyashka [42]
I don’t get it explain
5 0
3 years ago
An athlete whose mass is 87.0 kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constan
expeople1 [14]

Answer: X = 52,314.12 N

Explanation: Let X be the force the feet of the athlete exerts on the floor.

According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)

Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.

The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s

The acceleration of the barbell is gotten by using the equation of constant acceleration motion

S= ut + 1/2at²

But u = 0

S = 1/2at²

0.65 = 1/2 *a (1.3)²

0.65 = 1.69 * a/2

0.65 * 2 = 1.69 * a

a = 0.65 * 2/ 1.69

a = 0.77m/s²

According to newton's second law of motion

Resultant force = mass * acceleration

And resultant force in this case is

X - 52,200 = (87 + 61.22) * 0.77

X - 52,200 = 148.22 * 0.77

X - 52, 200 = 114.132

X = 114.132 + 52,200

X = 52,314.12 N

6 0
3 years ago
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