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mars1129 [50]
3 years ago
7

Where f is force, m is mass, v is velocity, and t is duration of impact, what other information do you need in order to determin

e the force of the car at the impact
Physics
1 answer:
Lelu [443]3 years ago
4 0

newton's second law helps here.

mass times star velocuty minus mass times end velocity divided by time = force.

short time ,,, hig force

long time lower force ... padding reduces shock ... impulse

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Which statements best characterize conductors? Check all that apply.
romanna [79]
It would be the first one and the third one
5 0
3 years ago
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A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest
yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}=v_{i}-gt (1)

Where:

  • v(i) is the initial velocity
  • v(f) is the final velocity
  • g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

0=30-9.81t

Solve it for t:

t=3.06\: s

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}^{2}=v_{i}^{2}-2gh (2)

0=v_{i}^{2}-2gh

We know that the initial velocity is 30 m/s.

0=30^{2}-2gh

Solving it for h we have:  

h=\frac{30^{2}}{2*9.81}

h=45.87 \: m

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

10=30-(9.81t)

So, the time elapsed to get 10 m/s is:

t_{upward}=2.04\: s

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

t_{upward}=2.04+t  

t=1.02\: s

This is the time when the ball has 10 m/s downward.          

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

t=2t_{upward}

t=2*3.06      

t=6.12\: s

The displacement will be zero after 6.12 s.  

e) Here we need to find the time when v(f) is 15 m/s

15=30-gt

t=\frac{15}{9.81}  

t=1.53\: s

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0
2 years ago
The spool has a mass of 50 kg and a radius of gyration of ko = 0.280 m. if the 20-kg block a is released from rest, determine th
Rzqust [24]
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis 
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is 
T = 1/2 I₀w² + 1/2mAVA²

= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation  is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
5 0
3 years ago
Which of the following is not a part of kinetic molecular theory? A. Particles are in constant, random motion. B. When particles
liraira [26]
The answer, I believe, is, "B. When particles collide, no energy is lost."
7 0
3 years ago
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