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4 years ago

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Physics

1 answer:

Paha777 [63]4 years ago

4 0

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

$P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s$

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

$P_1=P_2$

$0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2$

$v_2=1 m/s$

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

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Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) resp

Sophie [7]

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: $x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ , where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

x-values:

$x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}$ = $\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}$ = $\frac{81}{41}$ = 1.9756

y-values:

$y_{cm}$ = $\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}$ = $\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}$ = $\frac{-90}{41}$ = -2.1951

center of mass:

(1.9756, -2.1951)

6 0

3 years ago

Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary

balandron [24]

A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
$W=mg$
where
m is the mass
$g= \frac{GM}{r^2}$ is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, $g=9.81 m/s^2$ , so Gary's weight is
$W=mg=9.81 m$ (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is
$R=6370 km$
So the distance of Gary from the Earth's center is
$r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m$

The Earth's mass is $M=5.97 \cdot 10^{24} kg$ , so the gravitational acceleration is
$g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2$

Therefore, Gary's weight at 300 km above Earth's surface is
$W' = mg' = 8.95 m$ (2)

If we compare (1) and (2), we find that Gary's weight has changed by
$\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91$
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.

6 0

3 years ago

Estimate the number of Ping-Pong balls that can be packed into an average size room (without crushing them). Given that Ping-Pon

scoray [572]

Answer: 3,893,845.918 Ping-Pong balls

Explanation:

The volume of an average room is:

$V_{room}=(length)(width)(height)$ (1)

$V_{room}=(12 ft)(18 ft)(9 ft)=1944 ft^{3}$ (2)

Now let’s transform this $V_{room}$ to units of $cm^{3}$ , knowing $1 ft=30.48 cm$ :

$V_{room}=1944 ft^{3}\frac{{(30.48 cm)}^{3}}{1ft^{3}}=55,047,949.78 cm^{3}$ (3)

On the other hand, we have Ping-Pong balls with a radius $r=1.5 cm$ , and their volume is given by:

$V_{balls}=\frac{4}{3} \pi r^{3}$ (4)

$V_{balls}=\frac{4}{3} \pi (1.5 cm)^{3}$ (5)

$V_{balls}=14.137 cm^{3}$ (6)

Now, the number $n$ of Ping-Pong balls that can be packed into the room is:

$n=\frac{V_{room}}{V_{balls}}$ (7)

$n=\frac{55,047,949.78 cm^{3}}{14.137 cm^{3}}$ (8)

$n=3,893,845.918$ This is the number of Ping-Pong balls that can be packed into an average size room

7 0

3 years ago

Imagine tying a string to ball and twirling it around you.how is this similar to the moon orbiting earth?in this example what is

guapka [62]

The changing force is you swinging the STRING
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this is similar because it’s on a set path and you are the earth and the gravitational pull is from the earth which the string represents

5 0

4 years ago

Meghan explains to Zach the difference between commensalism, mutualism and parasitism. Which sentence(s) did she include in her

vlabodo [156]

C is the correct answer I’m sure

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4 years ago

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