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2 years ago

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A 0.180-kilogram cart traveling at 0.80 meter per second to the right collides with a 0.100-kilogram cart initially at rest. The

carts lock together upon collision. Calculate the final velocity of the carts.

1 answer:

My name is Ann [436]2 years ago

3 0

According to the <u>saving of the momentum</u> law, the total momentum <em>before </em>collision equals to the total momentum <em>after </em>collision

After collection, the 2 carts lock together, one body with a mass $m=m_{1}+m_{2}$ and velocity $v$

$(P_{1}+P_{2})_{before}=P_{after}$

from the definition of the momentum $P=mv$

$P_{1,before}=m_{1}v_{1,before}=0.180*0.80=0.144kg.m/s$

$P_{2,before}=m_{2}v_{1,after}=0.100*0=0kg.m/s$

thus

$P_{after}=(0.144+0)_{before}=0.144 kg.m/s$

we calculate the velocity

$v=\frac{P}{m}=\frac{0.144}{0.180+0.100}= 0.514m/s$

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For nitrogen feel like with its temperature must be within 12.78 Fahrenheit of -333.22 Fahrenheit which equation can be used to

photoshop1234 [79]

Answer:

The following equation can be used.

(32°F − 32) × 5/9=C

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2 years ago

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The gravity tractor is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to impact the Earth.

Talja [164]

Answer:

$F_g=461lb_f$

Explanation:

First calculate the mass of the asteroid. To do so, you need to find the volume and know the density of iron.

If r = d/2 = 645ft, then:

$V = \frac{4}{3} \pi r^3$

$V = 1.124*10^{9}ft^3$

So

$\delta_{iron}=m/V=491lb/ft^3$

$m=V*\delta=5.519*10^{11}lb$

Once you know both masses, you can calculate the force using Newton's universal law of gravitation:

$F_g=G\frac{m_1m_2}{d^2}$

Where G is the gravitational constant:

$G= 1.068846 * 10^{-9} ft^3 lb^{-1} s^{-2}$

$F_g=461lb_f$

4 0

2 years ago

13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?

sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

$Q = mc del T$

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

$Q = 20*C*80$

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

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2 years ago

(8%) Problem 9: Helium is a very important element for both industrial and research applications. In its gas form it can be used

exis [7]

Answer:

2046.37 kPa

Explanation:

Given:

Number of moles, n = 125

Temperature, T = 20° C = 20 + 273 = 293 K

Radius of the cylinder, r = 17 cm = 0.17 m

Height of the cylinder, h = 1.64 m

thus,

volume of the cylinder, V = πr²h

= π × 0.17² × 1.64

= 0.148 m³

Now,

From the ideal gas law

we have

PV = nRT

here,

P is the pressure

R is the ideal gas constant = 8.314 J / mol. K

thus,

P × 0.148 = 125 × 8.314 × 293

or

P × 0.148 = 304500.25

or

P = 2046372.64 Pa = 2046.37 kPa

6 0

2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

b)

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

2 years ago