All categories

2 years ago

You don't need to push off with a foot against the ground to start

Physics

1 answer:

Angelina_Jolie [31]2 years ago

4 0

Answer:

No, it does not violate Newton's first law

Explanation:

Newton's first law of motion states that, "A body will continue in its state of rest or uniform motion in a straight unless acted up by a force to make it act otherwise" This means a force is required to initiate movement and also required to halt it.

For a skate positioned to roll down a bank, Here, the force of gravity acting on the skate acting downward will make the skate roll without having to push off the skate with a foot. Because the position of the skate doesn't balance the Gravitational force acting on it. Hence. The Gravitational force is enough to set the skate in motion.

You might be interested in

1. Faça as transformações:

zhannawk [14.2K]

Answer:

$seconds (s) = hours (h) *3,600 ; h = \frac{s}{3,600} \\g = kg * 1,000; kg = \frac{g}{1,000} \\cm = \frac{m}{100};m = cm * 100$

1. a) 0.5 h = 1,800 s

h) 20 cm = 0.2 m

b) 2.0 h = 7,200 s

i) 5.0 kg = 5,000 g

c) 3.5 h = 12,600 s

j) 1.5 kg = 1,500 g

d) 1/4 h = 900 s

k) 450.0 g = 0.45 kg

e) 3.0 m = 300 cm

l) 20.0 g = 0.02 kg

f) 2.5 m = 250 cm

m) 500.0 g = 0.5 kg

g) 0.5 m = 500 mm

n) 1000.0 g = 1 kg

3 0

2 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

2 years ago

The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

$6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}$

$6.5\text{ Light-years}=1.9929065\text{ Parsecs}$

$6.5\text{ Light-years}\approx 1.99\text{ Parsecs}$

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

5 0

3 years ago

Please help with this!!!!!

34kurt

(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K) I hoped that helped

4 0

3 years ago

Write a simple rule that will tell a person how many water molecule will be lost while putting monosaccharides together to form

belka [17]

<span>For hydrolysis to monosaccharides, one molecule of a disaccharide needs only one molecule of water. C12H22O11 (sucrose) + H2O = C6H12O6 (glucose) + C6H12O6 (fructose) Structurally, a disaccharide molecule may be viewed as a product formed by the condensation of two molecules of monosaccharides with the elimination of a water molecule. So, only one H2O molecule is needed for the reverse process.</span>

3 0

2 years ago