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snow_tiger [21]
3 years ago
10

A 1600kg cannon fires a 5kg cannonball horizontally. The exit velocity of the cannonball is 80m/s and the barrel length is 2m. W

hat is the average acceleration of the cannonball before it leaves the barrel?
Physics
1 answer:
evablogger [386]3 years ago
7 0

Answer:

 a = 1600 m / s²

Explanation:

For this exercise we use the kinematics relations,

         v² = v₀² + 2 a x

where v₀ is the initial velocity of the bullet, which as part of rest is zero, for the distance (x) we can assume that the gases accelerate along the entire trajectory of the cannon x = 2m

         a = \frac{v^2}{2x}

let's calculate

         a = \frac{80^2}{2 \ 2}

         a = 1600 m / s²

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When a cup is placed on a table, which force prevents the cup from falling to the ground?
zhuklara [117]

Answer:

B. normal force

Explanation:

Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.

6 0
3 years ago
In a race, Usain Bolt accelerates at
jeka94

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

3 0
3 years ago
What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

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alexgriva [62]

Answer:

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