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GREYUIT [131]
2 years ago
13

A body starts moving from rest and attends the acceleration of 0.5m/s². calculate the velocity at the end of 3 minutes also find

the distance travelled by it during that time
​
Physics
1 answer:
puteri [66]2 years ago
4 0

Answer:

See below

Explanation:

3 minutes = 180 seconds

v f = at  =  .5 m/s^2 * 180 s =<u> 90 m/s </u>

<u />

Distance =  1/2 a t^2 = .5 * .5 * 180^2 =<u> 8100 m </u>

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3 years ago
Change the speed 0.200 cm/s to units kilometers per year
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5 0
3 years ago
A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal
givi [52]
I'm gonna have to assume the girl is on the right side and boy on left.
The net force is the sum of all forces on an object (includes negatives).
Let's say the force of the boy is variable <em>b</em>.  Use the formula F = ma.

<em>b </em>+ 3.5 = 0.2(2.5)

This is now simple algebra.  Solve to get that <em />the boty is exerting a force of -3N to the left.
5 0
3 years ago
Read 2 more answers
Will mark as brainliest if correct!!!!!
irakobra [83]

Refraction refers to C. the bending of light rays when they pass from one medium into another

Explanation:

Refraction is a phenomenon typical of wave. Refraction occurs when a wave travels through the boundary between two different mediums. When this occurs, the wave changes speed, wavelength and direction (but the frequency remains the same).

In particular, the direction of the refracted ray is determined by Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1 is the index of refraction of the 1st medium

n_2 is the index of refraction of the 2nd medium

\theta_1 is the angle of incidence, which is the angle between the direction of the incident wave and the normal to the boundary

\theta_2 is the angle of refraction, which is the angle between the direction of the refracted wave and the normal to the boundary

Therefore, the correct description of refraction is

C. the bending of light rays when they pass from one medium into another

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

8 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
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