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sveta [45]
3 years ago
5

You are a pirate working for dread pirate roberts. you are in charge of a cannon that exerts a force 10000 n on a cannon ball wh

ile the ball is in the barrel of the cannon. the length of the cannon barrel is 2.44 m and the cannon is aimed at a 45◦ angle from the
Physics
1 answer:
KATRIN_1 [288]3 years ago
7 0
<span>Answer: Let m = mass of cannon Then 10000 = ma a = 10000/m v^2 = u^2 + 2as v^2 = 0 + 2as 84^2 = 2(2.21)(10000/m) 84^2 m = 4.42(10000) m = 6.264172336 = 6.26 kg Part 2 Range = u^2sin(2x38)/g = 84^2sin(76)/9.8 = 698.6129229 = 698.6 m</span>
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Two dogs are pulling on a chew toy. One dog pulls the chew toy with 64 N [E] and
PIT_PIT [208]

Answer:

Eastward, at 11 m/s^2

Explanation:

64N-31N=unbalanced force of 33N

F=ma

33N=(3kg)a

a=11m/s^2 to the East

3 0
2 years ago
HELP
Ivanshal [37]

Answer:

The velocity is 60 km/hr.

Explanation:

<h3><u>Given:</u></h3>

Displacement (d) = 480 km = 48000 m

Time (t) = 8 Hours = 480 minute

Velocity (v) = ?

Now,

Velocity = Displacement ÷ Time

v = d/t

v = 480/8

v = 60 km/hr

Thus, The velocity is 60 km/hr.

<u>-TheUnknownScientist 72</u>

5 0
2 years ago
2. A hanging wind-chime on a calm day would have kinetic or potential energy?
rjkz [21]

Answer:

it would have potential energy

7 0
3 years ago
Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject
Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}

where

\mu_{sun} is gravitational parameter of sun =  1.32712 x 10^20 m^3 s^2.t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}

t = 12,105.96 sec

6 0
3 years ago
0/2 File Limit
slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

8 0
3 years ago
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