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Naya [18.7K]
3 years ago
6

All scientists try to base their conclusions on

Physics
2 answers:
Crank3 years ago
5 0
They base their conclusions off Data that they recorded when conducting experiments.
Dima020 [189]3 years ago
3 0
Mathematics and science
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An airplane is .68 Kilometers long. How many Millimeters long is the plane?
xxTIMURxx [149]

Answer:

680000

mark brainliest

4 0
2 years ago
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. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of
Rina8888 [55]

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

F_f = \mu W

where

\mu is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

\mu = 0.55 (coefficient of friction)

W=580 N (weight of the leaves)

Substituting, we find:

F=(0.55)(580)=319 N

Learn more about force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 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
4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two
Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=mV_{o}+MU_{o} (2)

p_{f}=(m+M)V_{f} (3)

m=1200 kg is the mass of the first car

V_{o}=20 m/s is the velocity of the first car, to the North

M=1400 kg is the mass of the second car

U_{o}=-22 m/s is the mass of the second car, to the South

V_{f} is the final velocity of both cars after the collision

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg} (6)

Finally:

V_{f}=-2.61 m/s (7) This is the resulting velocity of the wreckage, to the south

7 0
3 years ago
Ranboo <br><br> ⏚⟒⟒⌿<br><br> :) &lt;3 have a good day
bearhunter [10]
Ranboo oobnar have a good day
4 0
3 years ago
Read 2 more answers
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