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vesna_86 [32]
3 years ago
14

What is always true about the lateral magnification of an inverted image?

Physics
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

it is negative

..........

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A boat takes off from the dock at 2.5m/s and speeds up at 4.2m/s2 for 6.0 seconds how far has the boat traveled
suter [353]

Answer:

The answer is "91 m"

Explanation:

v_0 = 2.5 \ \frac{m}{s}\\\\a = 4.2\ \frac{m}{s^2}\\\\t = 6.0\ s\\\\\Delta x=?

Using formula:

\Delta x = v_0 t +\frac{1}{2} at^2\\\\

      = 2.5 \times 6.0  +\frac{1}{2}  \times  4.2 \times 6.0^2\\\\= 90.6\  m \approx 91\ m

5 0
3 years ago
Which of the following is true of children with chronic illness? a.) They are all eligible to recievie special education service
ss7ja [257]

Answer:

c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.

Explanation:

Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.

The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.

5 0
3 years ago
A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before contin
jeka94

a) See free-body diagram in attachment

b) Net force in the y-direction: F_y=mg+N[/tex]

c) The velocity at which the roller coaster will fall is [tex]v=\sqrt{gr}[/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

  • The weight of the roller coaster, acting downward, indicated by mg (where m is the mass of the roller coaster and g is the acceleration of gravity)
  • The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

F_y = mg + N

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

m\frac{v^2}{r}=mg+N

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

m\frac{v^2}{r}=mg+N

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

v=\sqrt{gr+\frac{Nr}{m}}

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

v=\sqrt{gr}

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

g=9.8 m/s^2

We said that the minimum velocity that the cart must have in order not to fall at the top is

v=\sqrt{gr}

And substituting, we find

v=\sqrt{(9.8)(30)}=17.1 m/s

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

mgh = \frac{1}{2}mv^2 + mg(2r)

where

h is the initial height at the starting point

(2r) is the height of the roller coaster at the top of the loop

We can re-arrange the equation making h the subject,

h=\frac{v^2}{2g}+2r

And substituting the minimum speed of the cart,

v=\sqrt{gr}

this becomes

h=r+2r=3r

And since r = 30 m, we find

h=3(30)=90 m

f)

In this case, 10% of the initial energy is lost during the motion of the roller coaster. We can rewrite the equation of the previous part as

0.90mgh = \frac{1}{2}mv^2 + mg(2r)

Because only 90% (0.90) of the initial energy is converted into useful energy (kinetic+potential) when the cart reaches the top of the loop.

Re-arranging the equation, this time we get

h=\frac{\frac{v^2}{2g}+2r}{0.90}

Again, by substituting v=\sqrt{gr}, we get

h=\frac{3r}{0.90}

And therefore, the new initial height must be

h=\frac{3(30)}{0.9}=100 m

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
The cheetah can reach a top speed of 114 km/h (71 mi/h). While chasing its prey in a short sprint, a cheetah starts from rest an
Mariulka [41]

Answer: a= 4.4ms-2

Displacement= 26.95 m

Explanation:

First, the speed in km/hr must be converted to m/s so that we can apply it in solving the question. The motion started from rest hence the initial velocity is 0m/s. The average displacement is also obtained from the equations of motion as shown in the image attached.

7 0
4 years ago
Read 2 more answers
01:28:29
Novosadov [1.4K]

Answer:

an inverse relationship

7 0
3 years ago
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