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2 years ago

an ice Puck travels 18 m in 3 S before it slides into the goal what is the speed of the traveling puck

Physics

2 answers:

USPshnik [31]2 years ago

6 0

Answer:

<h2>Given :-</h2>

Distance travelled (D) = 18 m
Time taken (T) = 3 second

Speed

<h2>Solution :-</h2>

As we know that

$\sf Speed = \dfrac{Distance}{Time}$

$\sf \: Speed = \dfrac{18}{3}$

Cancelling both by 3

$\sf \: Speed = 6 \: mps$

DiKsa [7]2 years ago

4 0

Answer:

$\boxed {\boxed {\sf 6 \ meters \ per \ second}}$

Explanation:

Speed can be found by dividing the distance by the time.

$s=\frac{d}{t}$

The distance is 18 meters and the time is 3 seconds.

$d= 18 \ m \\t= 3 \ s\\$

Substitute the values into the formula.

$s=\frac{18 \ m }{ 3 \ s }$

Divide.

$s= 6 \ m/s$

The speed of the puck is 6 meters per second.

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An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field

lawyer [7]

Answer:

a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

Em₀ = U = e DV

final point. Where they hit the target

Em_f = K = ½ m v2

energy is conserved

Em₀ = Em_f

e ΔV = ½ m v²

ΔV = $\frac{1}{2}$ mv²/e (1)

If the speed of light is c and this is 100% then 1% is

v = 1% c = c / 100

v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

ΔV = $\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }$

ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

$K_e = K_p$

K_p = ½ m v_e²

K_p = $\frac{1}{2}$ 9.1 10⁻³¹ (3 10⁶)²

K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

ΔV = Kp / M

ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

e ΔV = ½ M v²

v² = 2 e ΔV / M

v² = $\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }$

v² = 49,035 10⁸

v = 7 10⁴ m / s

Relation

v/c = $\frac{7 \ 10^4 }{ 3 \ 10^8}$

v/c= 2.33 10⁻⁴

8 0

3 years ago

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

$\Delta y = 0$

$0 = v_y t - \frac{1}{2}gt^2$

$T = \frac{2v_y}{g}$

$T = \frac{2(22.94)}{9.81} = 4.68 s$

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3 years ago

What is frost weddging

kiruha [24]

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3 years ago

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A 25 kg rock resting on the bottom of a lake must be moved from the paths of boats. The rock has a density of 2350 kg/m^3. What

algol13

Answer:

The force needed is the weight of the rock minus the buoyant force.

Explanation:

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3 years ago

What charge do electrons have? Negative Positive Neutral

hoa [83]

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3 years ago

an ice Puck travels 18 m in 3 S before it slides into the goal what is the speed of the traveling puck ​

an ice Puck travels 18 m in 3 S before it slides into the goal what is the speed of the traveling puck