Answer the question please

If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100

balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s = (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0

3 years ago

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)= $3.2 m/s^2$

Speed of car after 4.5 s

using equation of motion

v=u+at

$v=4.92+3.2\times 4.5=4.92+14.4$

v=19.32 m/s

Displacement of the car after 4.5 s

$v^2-u^2=2as$

$19.32^2-4.92^2=2\times 3.2\times s$

$349.05=2\times 3.2\times s$

s=54.54 m

4 0

3 years ago

Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm

Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining $\rho_A:$ Ethyl alcohol density and $\rho_G:$ Glycerin density , we can write:

$\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3$

Simplifying:

$\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)$

On the other hand:

$h_1 + h_2 = \Delta h + h_3$

Rearranging:

$h_1 - \Delta h = h_3 - h_2 (2)$

Replacing (2) in (1):

$\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)$

Rearranging:

$\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h$

Data:

$h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}$

$\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}} = \Delta h$

$7.5 cm = \Delta h$

7 0

3 years ago

Two spacecraft are both 10 million kilometers from a star. The total power output of the star is 4 x 1025 W. Spacecraft 1 has a

Ksenya-84 [330]

The concept of power is given by the relationship between intensity and area, that is to say that power is defined as

$P = A*I$

Our values are given under the condition of,

$r_1 = 18m$

$r_2 = m$

The power is proportional to the Area, and in turn, we know that the Area of a circle is the product between $\pi$ times the radius squared, therefore the power is proportional to the radius squared.

$\text{Power} \propto r^2$

For both panels we would have to

$\frac{\text{Power by panel 1}}{\text{Power by panel 2}} = \frac{r_1^2}{r_2^2}$

$\frac{P_1}{P_2} = (\frac{18}{6})^2$

$\frac{P_1}{P_2} = 9$

Therefore the correct option is option C.9

5 0

3 years ago

Read 2 more answers

The effect of itching power what’s the Independent variable and dependent and constants?

kenny6666 [7]

The independent variable is what type of powder is used, the dependent variable is how long itching lasts and the constant is the amount of powder used (50 grams)

7 0

3 years ago