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kondaur [170]
3 years ago
9

Find the instantaneous velocity at 1 s . can anyone help with c-h!!!

Physics
1 answer:
Lelu [443]3 years ago
7 0
Rise over run at 1 second
It’s the same slope from 0 to 2 seconds
10/2=5mps
As a note all time points between 0and 2 will have this instantaneous velocity

Instantaneous velocity at time 2 is 0
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A light wave encounters a partial physical barrier, such as a wall with a hole in it. What is MOST LIKELY to occur?
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Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
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n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
Frequency is measured in units called?
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3 years ago
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

4 0
3 years ago
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