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3 years ago

An object with a mass of 90 grams, moving at a constant velocity of 6 meters per second, has __________.

Physics

2 answers:

ale4655 [162]3 years ago

8 0

Answer:

a momentum of 540 grams per meters per second

Explanation:

son4ous [18]3 years ago

5 0

<span>A. a momentum of 540 grams per meters per second The momentum equation is p = mv. P is the variable for momentum, m is the variable for mass and v is the variable for velocity. To find the momentum, you must multiple the mass of the object by the velocity it is traveling by. The unit for momentum is the mass per meter per second.</span>

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Mr.maxwell, an english teacher, thinks all the sudents in the class who play sports are lazy

ValentinkaMS [17]

Slap em Bc they not supposed to me sleeping in her class and that’s on period

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3 years ago

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A 1500 kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h

Mandarinka [93]

Answer:

Incomplete question check attachment for diagram

Explanation:

Given that,

Mass of car

M = 1500kg

Enter curve at Point A with speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car was slow down at a constant rate till it gets to point C at speed of

Vc = 50km/r = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as given in the attachment is

S=200m

We want to find the total horizontal forces at point A, B and C exerted by the road on the tire

The constant tangential acceleration can be calculated using equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative because the car was decelerating

Since the car is slowing down at a constant rate, the tangential acceleration is equal at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can calculate the normal component of acceleration(centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration is calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is point of inflection

Then,

an = Vb²/p = Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Since tangential acceleration is constant, then, this is the tangential force at each point A, B and C

Now, normal force

Point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Then, the horizontal force acting at each point is

Using Vector of right angle triangle

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

6 0

3 years ago

The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) $v_{1} = 1.88\sqrt{3} \frac{m}{s}$

$v_{2} = 6.79 \frac{m}{s}$

Explanation:

Ball 1:

$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2} = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} = v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) + v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1} }{2cos(\theta)})^{2} = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) - v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0

3 years ago

ANSWER TRUE OR FALSE FOR EACH OF THESE.

Anestetic [448]

1.t
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4 years ago

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What's the climate change over the past 200 years

Mandarinka [93]

Answer:

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3 years ago

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