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RideAnS [48]
3 years ago
9

on earth a block is placed on a frictionless table. When a 50 North horizontal force is applied to the block, it accelerates at

4.0 m / s squared. Supposed to block and table are set up on the moon. When a horizontal force of 25 n is applied to the block what is the acceleration
Physics
1 answer:
melamori03 [73]3 years ago
5 0

As per Newton's II law we know that

F = ma

here

F = applied unbalanced force

m = mass of object

a = acceleration of object

now it is given that force F = 50 N North applied on block on earth due to which block will accelerate by 4 m/s^2

so here from above equation

50 = m* 4

m = \frac{50}{4} = 12.5 kg

Now we took another situation where block is placed on surface of moon and again force F = 25 N is applied on the block

So we will again use Newton's II law

F = ma

25 = 12.5 * a

a = \frac{25}{12.5}

a = 2 m/s^2

so block will accelerate on moon by acceleration 2 m/s^2

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For the following statements, indicate whether the statement is true or false and explain in words your reasoning.
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Answer:

a) If I increase the number of moving charged particles per cubic meter in a conductor, I will have to increase the cross sectional area of the conductor to get the same amount of current for the same electric field in the conductor.

FALSE

As we know that

I = neAv_d

so here if we increase the number of charge per unit volume then we need to decrease the area to get same amount of current through the wire.

b) The change in magnetic flux in a closed loop induces an EMF that opposes the change in magnetic flux.

TRUE

As per lenz law the direction of induced EMF is always opposite to the change in the flux due to which EMF is induced in the closed loop.

c) The magnetic force per unit length of two long, parallel, current-carrying conductors is repulsive if the currents are flowing in the opposite direction.

TRUE

When magnetic field of first wire will interact with other then it will exert force on it

so here the force will be on other wire such that the two wire will repel when current in the two wires is opposite in direction.

d) . The total magnetic flux through a closed surface is equal to μ0I enclosed

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magnetic flux is the number of field lines passing through a given area

So it is given as

B. A = 0

8 0
3 years ago
A similarity between radio waves and microwaves is that both are used for _____.
Cloud [144]

Answer:

(B) Sending Messages

Explanation:

  • A similarity between radio waves and microwaves is that both are used for sending messages and communication

Radio Waves and Microwaves have longer wavelengths and lower frequencies. Longer wavelengths than an x-ray which is in the electromagnetic spectrum and lower frequencies than gamma rays, which are also in the electromagnetic spectrum. They are also both electromagnetic radiation.

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4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.
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The answer is B.Jupiter

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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:

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#LearnwithBrainly

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