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Eva8 [605]
3 years ago
5

Which statement correctly defines the term scientific theory?

Physics
1 answer:
Sholpan [36]3 years ago
4 0

Answer:

B) an explanation for observable phenomena.

Explanation:

All theories have been stated by experiments and natural phenomenas.

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A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 35.0 min
Lubov Fominskaja [6]

Explanation:

Solution:

Let the time be

t1=35min = 0.58min

t2=10min=0.166min

t3=45min= 0.75min

t4=35min= 0.58min

let the velocities be

v1=100km/h

v2=55km/h

v3=35km/h

a. Determine the average speed for the trip. km/h

first we have to solve for the distance

S=s1+s2+s3

S= v1t1+v2t2+v3t3

S= 100*0.58+55*0.166+35*0.75

S=58+9.13+26.25

S=93.38km

V=S/t1+t2+t3+t4

V=93.38/0.58+0.166+0.75+0.58

V=93.38/2.076

V=44.98km/h

b. the distance is 93.38km

6 0
4 years ago
A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to
fredd [130]

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

4 0
2 years ago
A block slides along a frictionless surface and onto a slab with a rough surface. The slab has mass of 4 kg and the block has ma
maxonik [38]

Answer: the friction force on the small block at t equals 1 second is 2N

Explanation:

Given the data in the question;

from the slope in the graph provided, we will get the acceleration of the slab

At t = 1 seconds

Slope = acceleration = ( 1 - 0) / ( 2 - 0 ) = 1/2 = 0.5 m/s²

Force = ma = 4 × 0.5 = 2 N

so by Newton's third law

Force on block will be same which is 2N

Therefore the friction force on the small block at t equals 1 second is 2N

3 0
3 years ago
a small 20-kilogram canoe is floating downriver at a speed of 2m/s what is the canoes kinetic energy?
Vladimir [108]
KE = 1/2mv²
      = 1/2(20 kg)(2m/s²)
       = 1/2(20)(4)
       = 1/2 (80)
KE = 40 J

hope this helps :)



4 0
3 years ago
A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
  2. Particle's speed and direction of motion brainly.com/question/2814900
  3. About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0
3 years ago
Read 2 more answers
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