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4 years ago

5. Find the volume of the composite space figure to the nearest whole number.

1 answer:

6 0

The shape is missing but let's consider it a semi-cylinder attached to the rectangular prism.
Given:
radius = 4.5 mm
<span>Height = 11 mm </span>

<span>Volume of cylinder = (1/2)(pi)(4.5)^2(11) (the shape is divided into half)
V = 349.89 mm cubed
Volume of prism = L x W x H
= 9 x 11 x 6
= 594 mm cubed
Total volume of the composite shape = 111.375 + 594
= 943.89 mm cubed
Rounded answer = 944 mm cubed.</span>

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A 2-kg rock is thrown vertically upward at a speed of 3.2 m/s from the surface of the moon. If it returns to its starting point

worty [1.4K]

Answer:

1.6 m/s2

Explanation:

Let $g_m$ be the gravitational acceleration of the moon. We know that due to the law of energy conservation, kinetic energy (and speed) of the rock when being thrown upwards from the surface and when it returns to the surface is the same. Given that $g_m$ stays constant, we can conclude that the time it takes to reach its highest point, aka 0 velocity, is the same as the time it takes to fall down from that point to the surface, which is half of the total time, or 4 / 2 = 2 seconds.

So essentially it takes 2s to decelerate from 3.2 m/s to 0. We can use this information to calculate $g_m$

$g_m = \frac{\Delta v}{\Delta t} = \frac{0 - 3.2}{2} = \frac{-3.2}{2} = -1.6 m/s^2$

So the gravitational acceleration on the Moon is 1.6 m/s2

8 0

3 years ago

A car with a mass of 800 kg, moving at +15 m/s, crashes head on with a 1500 kg truck moving at -45.5 m/s in the opposite directi

Sloan [31]

Answer:

v = -24.5 m/s

Explanation:

Use conservation of momentum P = m*v. Unless there are external forces involved, total momentum must be conserved.

P: momentum

m: mass

v: velocity

1. calculate the total momentum before the collision:

$P=m_{car}v_{car} + m_{truck}v_{truck}$

2. the total momentum after the collision must not change:

$P_{before}=P_{after}$

3. after the collision the velocity of the car is the same as the velocity of the truck:

$v = v_{car} = v_{truck}$

4. combining 2. and 3.:

$P= (m_{car}+m_{truck})v$

5. this equation only has one unknown and can be solved for v:

$v = \frac{P}{m_{car}+m_{truck}} =\frac{m_{car}v_{car} + m_{truck}v_{truck}}{m_{car}+m_{truck}}$

$v=\frac{800*15 + 1500 * (-45.5)}{800 + 1500} = -24.5$

3 0

4 years ago

_______ was the first person to propose the idea of moving continents as a scientific hypothesis.

yan [13]

The answer is b alfred wegener

8 0

3 years ago

Read 2 more answers

a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv

Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

$J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m$

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

7 0

4 years ago

Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive

vazorg [7]

Answer: Yes, it can be a single, ordinary star.

Explanation: To determine a mass of a star, we use the orbital speed formula, given by: v = $\sqrt{\frac{GM}{R} }$ , where

v is the speed;

G is a constant: G = 6.67* $10^{-11}$ $\frac{m^{3} }{kg.s^{2} }$

M is mass of a massive object;

R is the distance between the object orbiting and the massive object;

The formula can be rewritten as:

$M = \frac{v^{2}.R }{G}$

First, we change R from light years to km:

1km=1.057* $10^{-13}$

R= $\frac{15}{2*1.057.10^{-13} }$

Calculating mass:

M = $\frac{2^{2}*10^{4}*14.2*10^{13} }{6.67*10^{-11} }$

M = 4.25* $10^{28}$ kg

A solar mass is the standard unit of mass. It is approximately 2* $10^{30}$ Kg and can be used for comparison: A single star cannot be more than 50 solar masses.

50 solar masses = 50*2* $10^{30}$ = $10^{32}$ kg

Comparing the mass of the object with this parameter, we have

$\frac{10^{32} }{4.25.10^{28} }$ = 0.235. $10^{4}$ = 2.35. $10^{3}$

From this, we know that 50 solar masses is greater than the small, massive object found. So, this object can be a single, ordinary star.

3 0

3 years ago