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Papessa [141]
3 years ago
13

What would happen if an automobile moving at a high speed suddenly comes to a stop? How does this relate to Newton's laws and wh

at factors could change the effect of the sudden stop on the passengers?
Physics
1 answer:
insens350 [35]3 years ago
6 0
If an automobile moving at high speed suddenly comes to a stop, you would have a large change in momentum. This relates to Newton's second law in the form F = delta p / delta t, where p is momentum (mv).
You could lessen the effect of the sudden stop on the passengers by changing the average force exerted on them. If you look at Newton's second law again, you can see that given some delta p, you can decrease F by increasing delta t. What this means is that if you increase the length of time over which the change in momentum occurs, you can decrease the average force exerted to obtain that change in momentum. This is the reason why landing on a soft cushion is preferable to landing on a concrete surface. The cushion gives way to any object falling on it while still providing some resistance (you don't stop as abruptly), so while your change in momentum is the same in both cases, you have a larger delta t in the case of the cushion.
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The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;
Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0
3 years ago
Given three different locations on Earth's surface, where will the weight of a person be greatest?
Feliz [49]

Answer:

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that  

F = \frac{Gm_{1} m_{2}  }{r^{2} }

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of G, m_{1} ,m_{2}, we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

4 0
3 years ago
You toss a walnut at a speed of 15.0 m/s at an angle of 50.0∘ above the horizontal. The launch point is on the roof of a buildin
kaheart [24]

Answer:

a.3.51s

b.33.8m

c. 9.64,-22.9

Explanation:

8 0
3 years ago
A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu
kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

\omega f = \omega o-\alpha*t

Solving for α:

\alpha=\frac{\omega o-\omega f}{t}

where ωo = 600*2*π/60;   ωf = 0;    t=10s

\alpha=6.283rad/s^2

The sum of torque is:

\tau=I*\alpha

\tau=M*R^2/2*\alpha

\tau=0.060 N.m

8 0
3 years ago
When the acceleration of a mass on a spring is zero, the velocity is at a
Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

F=kx

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

F=ma

where

m is the mass

a is the acceleration

So we can write the equation as

ma=kx (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const. (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a  maximum

When the velocity of a mass on a spring is zero, the acceleration is at a  maximum

6 0
3 years ago
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