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3 years ago

Phyics !!! Cannon ball question

Physics

1 answer:

natima [27]3 years ago

3 0

Answer:

Zero; no force is required to keep it going

Explanation:

Since the cannon ball is fired into frictionless space, there will be nothing to stop it, so it will keep going and going.

You might be interested in

If a force acting on an object is multiplied how is product called work?

Assoli18 [71]

Work is equal to the force applied to an object multiplied by the distance it had travelled. This is work as long as the force and the direction of the distance in going in the same direction.

It is work because you're applying a force on an object to move it.

For example, work is done when you push a box across the floor, but work is not done when you lift the box up and then walk it across the room. This is because the force on the box is upwards since you're lifting it and the direction of the distance is left or right so the directions are not the same.

6 0

3 years ago

Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the

murzikaleks [220]

Answer:

The function has a maximum in $x=3$

The maximum is:

$f(3) = 39$

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

$f(x)' = -4*2x + 24=0$

$-4*2x + 24=0$

$8x=24$

$x=3$

Now find the second derivative of the function and evaluate at x = 3.

If $f (3) ''< 0$ the function has a maximum

If $f (3) '' >0$ the function has a minimum

$f(x)''= 8$

Note that:

$f(3)''= -8$

the function has a maximum in $x=3$

The maximum is:

$f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39$

4 0

4 years ago

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

Scilla [17]

Answer:

d= 5.62 m

Explanation:

In order to have destructive interference, the path difference from the sources to the listener must be an odd multiple of half wavelengths, as follows:

d = (2n+1) * λ/2

In orfer to know which is the wavelength, we can use the relationship between propagation speed (in this case speed of sound), frequency and wavelength:

v= λ*f ⇒ λ = v/f = 344 m/s / 688 1/sec = 0.5 m

So, the path difference must be, at least, λ/2:

d = b-a = λ/2, where b is the distance to the speaker B, and a, the distance to the speaker A.

Applying Pithagorean Theorem, as the perpendicular distance d (which is our unknown) is the same for the triangles defined by the horizontal distance to the listener, and the straight line from the new position to the sources, we can write:

d² = a²- (3.0)²

d² = b²- (3.5)²

As the left sides are equal, so do right sides:

a² - (3.0)² = b² - (3.5)²

⇒ b² - a² = (3.5)² - (3.0)² = 3.25

We can replace (b²- a²) as follows:

b² - a² = (b+a)(b-a) = 3.25 (2)

We know that b-a = λ/2 = 0.25

Replacing in (2), we have:

b+a = 3.25 / 0.25 = 13 m

b-a = 0.25 m

Adding both sides:

2*b = 13.25 m ⇒ b= 13.25 /2 = 6.63 m

⇒ d² = (6.63)² - (3.5)² = 31.6 m²

⇒ d=√31.6 m² = 5.62 m

3 0

3 years ago

Which scenario is an example of a physical change?

klasskru [66]

Answer:

Lead is melted into a liquid to form pellets.

5 0

4 years ago

Read 2 more answers

The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i

Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

$T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1$

Block 2

$T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2$

Block 3

$- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3$

Solving Eq1,2,3 simultaneously

Divide 1 and 2

$\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4$

Put Eq 4 into Eq3

$T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5$

Put Eq 5 into Eq2 and solve for a

$a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6$

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

$T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\$

5 0

4 years ago