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LiRa [457]
3 years ago
8

PLSSSS help meeeeee lol

Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

B. As the temperature increases, the kinetic energy of the molecules increases.

Explanation:

When the temperature of an object increases, the kinetic energy of its particles increases, so the thermal energy of an object increases as its temperature increases.

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A.
Brut [27]

Answer:

\boxed{\sf Work \ done = 4 \ J}

Given:

Force = 8 N

Distance covered by the body = 50 cm = 0.5 m

Explanation:

Work Done = Force × Distance covered by the body

= 8 × 0.5

= 4 J

6 0
2 years ago
HELP ASP PLEASED AND THANKS NO LINKS PLEASED AND NO FILES
SOVA2 [1]

Answer:

just search up a ven-diagram and then try to draw it or trace it then use it for ur question

Explanation:

6 0
3 years ago
a vector is 253 m long and points in a 55.8 degree direction, what’s the y and x- component of the vector?
BartSMP [9]

Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

5 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
2 years ago
Compressing the air by squeezing the bottle was accompanied by a(n) ________ in the temperature of air inside the bottle
Georgia [21]
Increase in temprature
3 0
3 years ago
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