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nydimaria [60]
3 years ago
10

A scientific study is repeated by four different groups. Only two groups get the same results. Are the data reproducible?

Physics
1 answer:
tangare [24]3 years ago
4 0
No, one more group must get the same result as the other group. This will make it more reliable and reproducible.
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A stone is dropped from a cliff. What will be its speed when it was fallen 100 m?
Mars2501 [29]

Answer:

final velocity will be44.72m/s

Explanation:

HEIGHT=h=100m

vi=0m/s

vf=?

g=10m/s²

by using third equation of motion for bodies under gravity

2gh=(vf)²-(vi)²

evaluating the formula

2(10m/s²)(100m)=vf²-(0m/s)²

2000m²/s²=vf²

√2000m²/s²=√vf²

44.72m/s=vf

6 0
4 years ago
Read 2 more answers
A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
mash [69]

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0
4 years ago
A loaf of bread (volume 3100 cm3) with a density of 0.90 g/cm3 is crushed in the bottom of the grocery bag into a volume of 1240
andrew11 [14]

Answer:

2,25 g/cm3

Explanation:

Hi, you have to know one thing for this.. Density = mass/Volume,

When you have the loaf of bread with 3100 cm3 and a density of 0.90 g/cm3, the mass of that bread is 2790 g because of if you isolate the variable mass from the equation you get..  mass= density x volume

Later, have on account the mass never changes, so you crush the bread and the mass is the same.. so when you have the mashed bread.. you know that the mass is 2790 g and the volume of the bag is 1240 cm3, so you apply the main equation.... density=2790 g / 1240 cm3 , so density =  2,25 g/cm3

8 0
4 years ago
Calculate the kinetic energy in joules of a 1500 kg automobile moving at 19 m/s
Svetlanka [38]
Using
KE = ½mv² = ½×1500×19×19 = 270750 joules
5 0
3 years ago
Calculate the total surface area of cylinder whose base radius 2cm and height is 10cm​
murzikaleks [220]

Answer:

Answer :-  150.79cm2

Explanation:

                                               A=2πrh+2πr2

                                                 2·π·2·10+2·π·22

                                                    150.79645cm²

5 0
3 years ago
Read 2 more answers
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