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kolbaska11 [484]
2 years ago
9

Swinging a golf club toward a golf ball and hitting it off the tee.

Physics
2 answers:
Vlad1618 [11]2 years ago
7 0

Answer:

sucks cocka-doodle-doooooooo

Explanation:

he likes it jsjsjsjsjsjsjjjsnsns

belka [17]2 years ago
3 0
Yes okay hello bye lolol
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2) A car travels 5 miles north, and then 10 miles south. What is the person's DISTANCE and
Liono4ka [1.6K]

Answer:

distance traveled is 15 mi

displacement is 5 mi

Explanation:

Distance takes time into account and adds up all the tiny displacements during the entire period of the trip.

Displacement ignores time and looks only at the change in position from the starting point to the ending point.

6 0
3 years ago
Which statement accurately represents the arrangement of electrons in Bohr’s atomic model?
noname [10]

Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy

Explanation:

The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.

It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.

Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.

8 0
3 years ago
A 5.6 g marble is fired vertically upward using a spring gun. The spring must be compressed 6.4 cm if the marble is to just reac
N76 [4]

Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy

Explanation: i took the test and got a 100%

5 0
2 years ago
The coefficients of static and kinetic frictions for plastic on wood are 0.50 and 0.40,respectively. How much horizontal force w
IrinaK [193]

The horizontal force needed to start the calculator moving from rest is 1.5 N

What is Kinetic friction?

It is defined as a force that acts between moving surfaces.

The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.

Here,

weight of calculator, N = 3 N

The coefficients of static frictions, µ (static) = 0.50

The coefficients of kinetic frictions, µ (kinetic) = 0.40

Now,

The horizontal force required = The static friction force

F = µ (static) * weight of calculator

F = 0.50 * 3.0

F = 1.5 N

Hence,

The horizontal force needed to start the calculator moving from rest is 1.5 N

Learn more about   horizontal force here:

<u>brainly.com/question/21481680</u>

<u />

#SPJ4

5 0
2 years ago
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

8 0
1 year ago
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