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postnew [5]
4 years ago
8

1. Kinetic energy is the energy of _____.

Physics
1 answer:
lorasvet [3.4K]4 years ago
7 0

Explanation:

kinetic energy is energy that it possesses due to its motion.

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Can someone answer these
LenKa [72]

Four

Sometimes I think the creators of problems out to drawn and quartered. 60 g does not mean 60 grams. It means 60 * the acceleration due to gravity.

So the question really reads. The acceleration delivered by the air bag is 60 times that of a normal gravitational. This acceleration is delivered to the person where his mass is putting up a whole lot of resistance because he and his 75 kg are moving forward with the impact of the car. The 36 msec. has nothing to do with the problem.

The Force of the Air Bag is mass * a

F_airbag = mass * acceleration = 75 kg * 60 * 9.81 mass * acceleration = 44145 newtons

The answer is 4.41 * 10^4

Answer C

Five

This problem is governed by one formula that you sort of have to get out of your hat -- a piece of magic if you will.

Fg - Bf = m * a

Fg = the Force of gravity

Bf = the braking force

The mass of the rocket is derived from its weight

The acceleration is derived from one of your big 4 equations.

m of the rocket = 75600 / 9.81 = 7706 kg

The acceleration =

vi = 1 km/s = 1000 m/s

vf = 0

t = 2 minute * 60 sec/ min = 120 seconds

a = (vf - vi)/t = (0 - 1000 m/s) / 120 sec

a = - 8.333 m/s^2 The minus sign makes perfect sense. Remember the rocket is slowing down

The net downward force = mass * acceleration = - 7706 kg * - 8.333 m/s^2

The net force = - 64217 N

So going back to the problem's equation we have

Gravitational force - Braking Force = Net Force

Gravitational Force = 75600

Net Force = - 64217

Bracking force = ?

75600 - Bracking force = - 64217  Subtract 75600 from both sides

- Bracking force = - 64217 - 75600

- Braking force = - 139817

Braking force = 139817 N = 1.398 * 10^5 N

Braking Force = 1.4 * 10^5

Answer: Last One.

Six

The first thing you should do is derive a general formula for this problem.

The force pulling both masses down is M*g where g is the acceleration due to gravity.

The formula for this problem is

Mg = (m + M) * a

Now you need to solve for a

a =  [M/(M + m) ] * g

Look what is happening. is a smaller or larger than g? This is a question you should really pay attention to. If it was larger, everyone would have this system in their basement because you'd get more energy output than you put in. Something for nothing is always appealing.

So what's the answer? (I get to ask it. No one posing the question ever should).

A

A is incorrect. M never goes away. The acceleration may get very tiny, but there always is some acceleration.

B must be true. It is just what I finished saying about A

C Who said anything about velocity? It's a red herring. If the velocity became 0 the acceleration would have to turn minus. This answer sounds good, but sounds good doesn't make it right. C is wrong.

D The acceleration does not remain constant no matter what. The answer to A still applies. So D is wrong.

4 0
3 years ago
A box is 5 cm high, 4 cm wide, and 9 cm long. What is the
FromTheMoon [43]

Answer:

18

Explanation:

  1. 5+4=9+9=<em>18</em><em> </em><em>IS</em><em> </em><em>THE</em><em> </em><em>VOLUME</em><em> </em><em>OF</em><em> </em><em>CUBIC</em><em> </em><em>CENTIMETRES</em><em> </em>
6 0
3 years ago
A child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0.
kolezko [41]
In a problem where a child is danger form drowning from a river who has a current of 3.1km/hr to east and the child is 0.6km fro the shore and the upstream is 2,5km from the dock. So base on the question the boat with a speed of 24.8 km/hr is 1.9 km because the child is 0.6 km off the dock so 2.5 minus 0.6
4 0
3 years ago
show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t
timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

              (b) r_{p}= 9.95 x 10⁻¹ m

              (c) ratio = 1800

Explanation: (a) <u>Potential</u> <u>difference</u> is defined as the energy a charged particle has between two points in a circuit. It is calculated as

\Delta V=\frac{pe}{q}

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = \frac{1}{2}m.v^{2}

so

V.q=\frac{1}{2} m.v^{2}

V=\frac{m.v^{2}}{2q}

Calculating:

V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}

V = -109.44 x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is -109.44 x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

F_{mag} = F_{c}

qvB=\frac{mv^{2}}{r}

r=\frac{mv}{qB}

The radius of the curvature, for a proton, will be:

r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}

r = 54.33 x 10⁻⁵m

ratio = \frac{9.95.10^{-1}}{54.33.10^{-5}}

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

5 0
4 years ago
PLZ HELP
Sati [7]

Answer:

the index of refraction of the second medium is lower

Explanation:

take an exaple of a light ray from air to water that is optically denser the ray is refracted to the normal thus lowering its index of refraction

8 0
3 years ago
Read 2 more answers
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