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netineya [11]
4 years ago
13

Scientists know how groups of animals changed over time by studying information contained in the ______________.

Physics
2 answers:
Sedbober [7]4 years ago
5 0
The answer would be ROCKS because rocks contain fossils and the fossils can reveal a lot of the past life of extinct animals.
Arisa [49]4 years ago
3 0
Fossil succession is the blank answer.
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In Figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 N and the distance it acts over is 0.
pshichka [43]
I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
4 0
3 years ago
A force of 20 N is executed to raise a rock weighing 30 N. What is the actual mechanical advantage?
xxMikexx [17]
I am absolutely sure its 1.5
5 0
4 years ago
Read 2 more answers
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the
KiRa [710]

Answer:

The maximum amount of meat that the butcher can sell is  3\frac{1}{12}\:lb

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\

Best Regards!

8 0
3 years ago
The interest rate that the Fed charges banks that borrow money from it is called A. the discount rate. B. the federal rate. C. t
son4ous [18]
The answer is A. the discount rate.
7 0
3 years ago
Read 2 more answers
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
3 years ago
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