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2 years ago

A sled sliding on a flat,icy surface with a constant velocity is best described by

1 answer:

5 0

The first one is actually Newton's First Law of Motion.

Newton's First Law of Motion is commonly stated as "an object in motion will remain in motion unless acted upon by an outside force." This sled is moving across the ice at the same constant speed and velocity meaning it will won't stop unless an outside force acts upon it. An example being someone coming and grabbing the sled or maybe the sled slips off of the ice and finds itself stuck in a pile of snow. Similarly, if the sled was not moving, it would not move unless acted upon again by an outside force.

You believed it to be Newton's Third Law, but for future reference, his third law is more about every action having an equal opposite reaction. This wouldn't explain what is currently happening with the sled. A helpful tip into figuring out where to start in figuring out the answer I got is in the example you gave it literally states "with a constant velocity" which also ends up being in the description of Newton's First Law. Though, don't rely on that because the answer could have had a made up definition.

Unfortunately, I do not know the answer to your other question. I apologize. Hope this helped.

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A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

2 years ago

A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?

omeli [17]

The ball hitting the bat

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2 years ago

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Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio

Pachacha [2.7K]

First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N

Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>

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2 years ago

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A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial

puteri [66]

Answer:

A) vertically upward

Explanation:

Since the tyre is rotating with uniform angular speed and moving with constant linear speed

So as soon as a small stone is stuck into the groove of the tyre the speed of the stone is same as that of the tyre

so now we can say that stone will start revolving with the tyre of the car at constant angular speed and moving with uniform speed also

so here just after that the tangential acceleration of the stone must be zero while radial acceleration must be towards the center of the tyre given as

$a_c = \omega^2 R$

so we will have direction of net acceleration is towards its center so correct answer will be

A) vertically upward

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2 years ago

A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A seco

Roman55 [17]

Answer:

Explanation:

Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .

magnetic field in a torroid B = μ₀ n I , n is number of turns per unit length and I is current .

B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)

flux in the secondary coil

= B x area of face of coil x no of turns of secondary

= 4π x 10⁻⁷ x (1000 /2π x .16 ) .25 x 10⁻⁴ x 750

= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹

2343.75 x 10⁻⁸

= 23.43 x 0⁻⁶ H.

6 0

2 years ago

Read 2 more answers