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marissa [1.9K]
3 years ago
13

This is for physical science. Can someone help me?

Physics
2 answers:
Radda [10]3 years ago
6 0
A. 320 g
B. 160 g
C. 80 g
D. 40 g
romanna [79]3 years ago
5 0
You can use these steps for most if not all of the problems

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David rowed a boat upstream for three miles and then returned to point he started from. The entire journey took four hours. The
frez [133]
Upstream speed = S - 1
Downstream speed = S + 1

Average speed = total distance / total time

Average speed = (S - 1) + (S + 1) / 2
= S

S = 6 miles / 4 hours
S = 1.5 miles per hour
4 0
3 years ago
Which statement describes the endothermic reaction by this graph?
alexgriva [62]

Answer:

B

endothermic: heat taking in

exothermic: heat given out

8 0
3 years ago
Read 2 more answers
What are earths two internal sources of heat energy?
CaHeK987 [17]

the radiogenic heat produced by the radioactive decay of isotopes in the mantle and crust, and the primordial heat left over from the formation of the Earth.

3 0
3 years ago
The train took 2 h 30 min to travel 3/5 of a journey at an
tresset_1 [31]

Answer:

100 km/h

Explanation:

96km•2.5=240km

240÷3/5=240•5/3=400km

400-240=160km

4.1-2.5=1.6h

160÷1.6=100km/h

6 0
3 years ago
QuestionDetails:
sleet_krkn [62]

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

r = \sqrt{x^2+R^2}

Then the potential energy is

U = \frac{-GMm}{\sqrt{x^2+R^2}}

PART A) The force is excepted to be along x-axis.

Therefore we take a derivative of U with respect to x.

F = -\frac{dU}{dx}

F = -\frac{d}{dx}(GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}))

F = \frac{GMmx}{(x^2+R^2)^{3/2}}

This expression is the resultant magnitude of the Force F.

PART B) The magnitude of loss in potential energy as the particle falls to the center

U = GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})

According to conservation of energy,

\frac{1}{2}mv^2 = GMm (\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})

\therefore v = \sqrt{2GM(\frac{1}{R}-\frac{1}{x^2+R^2})}

7 0
3 years ago
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