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3 years ago

6

2. ____________________________ can cause a stationary object to start moving or a moving object to change its speed or directio

n or both.

1 answer:

Vikentia [17]3 years ago

7 0

As per Newton's law of inertia we can say every object will move in its state of motion either in the state of rest or will move with same constant velocity until some unbalanced external force will act upon it

so here we have to fill the space which says that stationary object can start or moving object will change its direction only in which case

so here the correct answer should be Unbalanced Force condition we will obtain such situation.

so we have

<u>Unbalance force</u> can cause a stationary object to start moving or a moving object to change its speed or direction or both.

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As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.1 s. If the train's acceleration remains constant

vodomira [7]

Acceleration is the rate of change of the velocity of a moving object. To determine the speed the object has at a certain time, we need to determine the acceleration first by dividing the speed with the time given. We do as follows:

acceleration = 4.9 m/s / 5.1 s = 0.96 m/s^2

To determine the velocity after 7.1 s, we multiply 7.1 s to the acceleration.

speed or velocity = 0.96 x 7.1 = 6.52 m/s

3 0

4 years ago

Read 2 more answers

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

6 0

4 years ago

A 500 kg rollercoaster car starts from rest at the top of a 10.0 m tall hill. it then travels down the track and up a loop. the

malfutka [58]

Speed of the roller coaster at the top of the loop= 7.67 m/s

Explanation:

using the law of conservation of energy

KEi + PEi= KEf + PEf

KEi= kinetic energy at the top of the hill=0 because the car is at rest there.

PEi= potential energy at the top of the hill

PEf= potential energy at the top of the loop

KEf= kinetic energy at the top of the loop

Also kinetic energy= 1/ 2m v² and potential energy= mgh

m= mass

h= height

v= velocity

so 0+ mghi = 1/2mv² + mg h

500 (9.8)(10)+ 1/2 (500) v²= 500 ( 9.8) (7)

49000+250 v²= 34300

250v²= 14700

v²=58.8

v=7.67 m/s

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3 years ago

Which of the following properties of materials does NOT affect resistance?

Shalnov [3]

Answer:

B thickness

Explanation:

thickness

4 0

3 years ago