What type of friction prevents a pile of rocks from falling apart?

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0

2 years ago

A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the

Shkiper50 [21]

Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

$K.E=\frac{1}{2}mv^2$

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg (1 kg = 1000 g)

v = velocity

Now put all the given values in the above formula, we get:

$K.E=\frac{1}{2}mv^2$

$6.8J=\frac{1}{2}\times 0.046kg\times v^2$

$6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2$

$v=17.19m/s\approx 17m/s$

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s

3 0

3 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

2 years ago

What is the potential energy of a 2 kg ball 15 m in the air?

oee [108]

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 2 × 10 × 15

We have the final answer as

Hope this helps you

7 0

3 years ago

__________ is the study of things continually moving or at rest.

bogdanovich [222]

Force

Explanation:

so force is your answer see why b/c

4 0

3 years ago