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Yuliya22 [10]
2 years ago
3

Think of a skateboarder riding down a hill. Once the skateboarder reaches the bottom of the hill, if she does not manually add m

otion, the skateboarder will slow until she stops. Why does the skateboarder eventually stop?
Physics
2 answers:
kicyunya [14]2 years ago
6 0
It will eventually stop because it does not have no energy to move it was just the air moving it at first . hope it helps
GarryVolchara [31]2 years ago
3 0

Answer:

The skateboarder eventually stop due to frictional force.

Explanation:

The Newton's first law of motion or law of inertia says that an object will remain in its previous state until and unless the unbalanced force acts on it.

When the skateboarder reaches the bottom of the hill, the skateboarder will slow until she stops. It slows down due to frictional force which is acting in opposite direction. It is a contact force and it acts between two surfaces.

Hence, the frictional force will act between the ground and the skateboard which reduces the speed of the skateboarder.

You might be interested in
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
The expressions for e/m and the relative error of e/m due to all of the parameters measured:
bija089 [108]

Answer:

Term 1 = (0.616 × 10⁻⁵)

Term 2 = (7.24 × 10⁻⁵)

Term 3 = (174 × 10⁻⁵)

Term 4 = (317 × 10⁻⁵)

(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) to the appropriate significant figures.

Explanation:

(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²

mean measurements

Voltage, V = (403 ± 1) V,

σᵥ = 1 V, V = 403 V

Current, I = (2.35 ± 0.01) A

σᵢ = 0.01 A, I = 2.35 A

Coils radius, R = (14.4 ± 0.3) cm

σʀ = 0.3 cm, R = 14.4 cm

Curvature of the electron trajectory, r = (7.1 ± 0.2) cm.

σᵣ = 0.2 cm, r = 7.1 cm

Term 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)

Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)

Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)

Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)

The relative value of the e/m ratio is a sum of all the calculated terms.

(σ ₑ/ₘ) / (e/m)

= (0.616 + 7.24 + 174 + 317) × 10⁻⁵

= (498.856 × 10⁻⁵)

= (499 × 10⁻⁵) to the appropriate significant figures.

Hope this Helps!!!

6 0
2 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The
Eddi Din [679]

Answer:

(a) Steel rod: 1.1 * 10^{-4}

    Copper rod: 1.88 * 10^{-4}

(b) Steel rod: 8.3 * 10^{-5} m

Copper rod: 1.41 * 10^{-4} m

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = \frac{1}{Y} * \frac{F}{A}

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y =  200 000 000 000 N/m^{2}

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

∴ Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}

For the copper rod:

Y =  120 000 000 000 N/m²

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain =  = 1.88 * 10^{-4}

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = 1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m

For the copper rod:

Elongation = 1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m

6 0
2 years ago
A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height
Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

v_x = 23.2 m/s

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

v_y(t)= v_{y0} -gt (1)

where

v_{y0}=0 is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s

Substituting t into (1) we find the final vertical velocity

v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}

6 0
2 years ago
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