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AlekseyPX
4 years ago
5

Kriste walks once around a large circle at a constant speed.

Physics
1 answer:
asambeis [7]4 years ago
4 0

Answer:

This question is incomplete

Explanation:

This question appears incomplete because of the absence of the value for the constant speed. However, there is a striking similarity between speed and velocity and the major difference between the two "quantities" is that speed is a scalar quantity while velocity is a vector quantity.

A scalar quantity (such as speed) is one that has a magnitude but no direction while a vector quantity (such as velocity) is one that has both magnitude and direction. Hence, <u>velocity can be described as the speed in a particular direction (around a large circle as suggested in the question)</u>.

From the description above, <u>it can be deduced that Kriste's average velocity is the same as her constant speed as she walks around the large circle</u>.

NOTE: The unit for both speed and velocity here will be in m/s

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A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const
Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

0= u^2-2as_{1}

s_{1}=\dfrac{u^2}{2a}....(I)

Using newton law

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{1}=\dfrac{8}{F}

Now, For second object

Using equation of motion

v^2=u^2+2as

Put the value in the equation

0= u^2-2as_{2}

s_{2}=\dfrac{u^2}{2a}....(I)

Using newton law

F = ma

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{2}=\dfrac{8}{F}

Hence, Both objects travel the same distance.

6 0
3 years ago
10/12
grandymaker [24]
25mph I hope this helps and sorry it took so long
7 0
3 years ago
A ball is thrown upwards at 20m/s what is its displacement from its starting position at 7 seconds
Gre4nikov [31]
Two hundred 432000 was broke down into aalll these parts
3 0
4 years ago
Tracy makes a list of energy transformations that occur as an engine runs to power a car. 1. chemical energy transforms into the
alex41 [277]

Transformation 3 should Tracy removed from the list. Chemical energy transforms into gravitational potential energy.

<h3> What is the law of conservation of energy?</h3>

According to the Law of Conservation of Energy, energy can neither be created nor destroyed, but it can be transferred from one form to another.

The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

Transformations that occur as an engine runs to power a car is;

1. Chemical energy of fuel transforms into thermal energy.

2. Thermal energy transforms into electrical energy used to lighten the bulb in a car.

3. Thermal energy transforms into kinetic energy to help to run the car.

Chemical energy transforms into gravitational potential energy is the incorrect energy transformation.

Hence, transformation 3 should Tracy removed from the list.

To learn more about the law of conservation of energy, refer to brainly.com/question/2137260.

#SPJ1

8 0
2 years ago
Although the evidence is weak, there has been concern in recent years over possible health effects from the magnetic fields gene
Natasha2012 [34]

Answer:

A. B = 6.36 * 10^{-10} T

B. P ≈ 0

Explanation:

In order to calculate the magnetic field strength we have to use the magnetic field strength of a straight wire.

B = \frac{mi* I}{2\pi *d} (eq. I)

B = magnetic field strength at distance d

I = current (A)

mi = represented by the greek letter μ, represents the permeability of the free space, which is: 4 × π 10^(-7) T m/A

d = distance from the wire

By replacing the values in eq I, we have the following:

B = \frac{4\pi  10^{-7} T  m  A^{-1}  200 A}{2\pi *20 m}\\\\B = 6.36 * 10^{-10}  T\\ (eq II)

The earth magnetic field in the surface variates from 25 to 65 microteslas. Thus:

P = Percentage from the wires/percentage of the earth

P = \frac{6.36 * 10^{-10}T}{65* 10^{-3} T}\\ ∵ B ∴

P ≈ 0

5 0
4 years ago
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