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2 years ago

15

An athlete is running at a constant velocity with a javelin held in his right hand. The force he is applying on the javelin to c

arry it is 5.0 newtons. If he covers a distance of 10 meters, what work has he done on the javelin?

2 answers:

Helen [10]2 years ago

8 0

Answer:

50J

Explanation:

Work done=Fd

=5×10

=50J

Anastaziya [24]2 years ago

7 0

50J
Equation= fd
= 5x10
=50

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The voyage of the H.M.S what important because

yaroslaw [1]

C. Physicalist chemist and biologist conducted experience in gather collection that changed existing scientific knowledge about the oceans

Explanation:

The voyage of Her Majesty's Ship was highly important because it furnished scientists with rich evidence about a lot of phenomenon under study and this changed existing scientific knowledge about the ocean.

H.M.S II was an expedition to study and map the ocean floor using some sophisticated and newer devices.
It was through this expedition that the theory of plate tectonics was modified and became a theory
The bathymetry of the ocean floor was also carried out.
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3 years ago

You work for NASA and an engineer. Your supervisor is thinking of building a space station instead of a

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3 years ago

How is a magnetic field generated?

alekssr [168]

It would be B I think

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3 years ago

A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j

fenix001 [56]

Answer:

$F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k$

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

$\vec F_e = q\vec E$

$\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3$

$\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j$

Now force on moving charge due to magnetic field is given as

$\vec F_b = q(\vec v \times \vec B)$

$\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )$

$\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i$

$\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k$

Now net force due to both

$F = F_e + F_b$

$F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k$

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3 years ago

21. A train went 300 miles from city X to city Y at an average speed of 80 mph. At what speed did it travel on the way back if i

patriot [66]

Answer:

c

Explanation:

133/3 mph

5 0

2 years ago

Read 2 more answers