what are the laws of newton
Answer:
Explanation:
These are the laws of Newton
Answer:
the first law, an object will not change its motion unless a force acts upon it. the 2nd one, the force of an object is equal to its mass times it acceleration. the 3rd one is when 2 objects interact, they apply forces to each other of equal magnitude and opposite direction.
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will ________ because the toy car ____
Answer:
If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration.
Explanation:
I hope this helped
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car gains more potential energy at the higher starting point.
When a toy car moves along a sloped racetrack, it converts its potential energy (due to its height above the ground) into kinetic energy (energy of motion). The higher the starting height of the racetrack, the more potential energy the toy car possesses initially.
As the toy car moves down the sloped track, it will accelerate due to the force of gravity. The potential energy is converted into kinetic energy, and the car's speed increases. According to the law of conservation of energy, the total mechanical energy (sum of potential and kinetic energy) remains constant as long as no external forces, such as friction, act on the car.
Therefore, if the starting height of the racetrack is increased, the toy car will have more potential energy to start with. As it moves down the track, it will convert this increased potential energy into kinetic energy, resulting in a higher speed compared to when it starts from a lower height.
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A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coefficient of friction was estimated to be 0.38. a) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.b) Why does the car's mass not matter?
Answer:
a. V = 19.1m/s
b. The mass of the car does not matter
Explanation:
A.
KE = 1/2mv² = fd --------(1)
Fd = umgd ---------(2)
Therefore,
1/2mv² = umgd ---------(3)
M will cancel itself out from both sides of equation 3.
Then we will have:
1/2v² = ugd
Then we cross multiply to make v² the subject of the formula
V² = 2ugd
V = √2ugd -------(4)
U = 0.38
g = 9.81
d = 98
When we input these values into equation 4, we will have:
V = √2x0.38x9.81x98
V = √730.6488
V = 27.03m/s
B.
The mass of the car does not actually matter as the mass was cancelled out on the both sides of equation 3