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liberstina [14]

2 years ago

15

5. A 1-kg car and a 2-kg car are both released from the top of the same hill and roll down a frictionless track. At the bottom o

f the hill, which car will have the greater
speed? Which car will have the greater Kinetic energy?

O A The 1-kg car will have greater speed, and the 2-kg car will have greater kinetic energy

OB. The 2 kg car car will have greater speed, and the 7-kg car will have greater kinetic energy

OC. The cars will have equal speeds and equal kinetic energies.

OD. The cars will have equal speeds and the 2 kg car will have greater kinetic energy

1 answer:

grigory [225]2 years ago

7 0

The cars will have equal speeds and the 2 kg car will have greater kinetic energy.

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A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>

3 0

2 years ago

The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1

antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

3 0

2 years ago

A proton (1.6726 ? 10-27 kg) and a neutron (1.6749 ? 10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or "

Alexus [3.1K]

Answer:

Explanation:

The mass of the deuteron = mass of the proton + mass of the neutron + mass equivalent of the energy of 2.2 Mev evolved.

I amu = 931 Mev

2.2 Mev = 2.2 / 931 amu

= ( 2.2 / 931 )x 1.6726 x 10⁻²⁷

= .00395 x 10⁻²⁷

The mass of the deuteron =( 1.6726 + 1.6749 + .00395)x 10⁻²⁷ kg

= 3.35145 x 10⁻²⁷ kg

b ) Momentum of gamma ray

= h / λ ( h is plank's constant and λ is wavelength of gamma ray )

= hυ / υλ ( υ is frequency of gamma ray )

= E / c ( E is energy of photon and c is velocity o light )

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹ J / 3 x 10⁸

= 1.173 x 10⁻²¹ Kg m /s

This will be the momentum of deuteron also

Kinetic energy

= p² / 2m ( p is momentum and m is mass of deuteron )

= ( 1.173 x 10⁻²¹ )² / ( 2 x 3.35145 x 10⁻²⁷)

= 1.376 x ⁻¹⁵ J

Energy of gamma ray

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 3.52 x 10⁻¹³ J

So kinetic energy of deuteron is smaller than energy of gamma ray photon .

5 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

Volgvan

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

<em>Lets explain how to solve the problem</em>

- The speed ramp has a length of 121 m and is moving at a speed of

2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

the ground

<em>Lets find your speed on the ground</em>

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

8 0

2 years ago