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ra1l [238]
3 years ago
6

One strategy in a snowball fight is to throw

Physics
1 answer:
faltersainse [42]3 years ago
7 0
I can solve your problem. Send me WA +524774333992
You might be interested in
True or false Weight is constant every place in the universe.
ycow [4]
False. it's depend on g -constant.

6 0
4 years ago
Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner t=\frac{295.6}{4.65}=63.56 s

time required by first runner to reach finish line=\frac{250}{3.1}=80.64 s

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0
4 years ago
Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
3 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0
3 years ago
.A particle starts on the origin. It is pushed back to -5.7 m in 2.1 s. Then it is pushed
harina [27]

Answer: The average velocity is -0.965m/s

Explanation: The first step is to calculate the two velocities is both directions. A velocity is a distance per unit time.

V=d/ t

=-5.7/2.1

=-2.7m/s

For the other direction the velocity is

V=7.3/9.5

=0.77m/s

The average velocity the add the velocities and divide them by 2.

V=-2.7+0.77/2

V= 0.965m/s

5 0
3 years ago
Read 2 more answers
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