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3 years ago

What best describes the dropping height of a ball that bounced back up to a height of 45 centimeters?

Physics

2 answers:

damaskus [11]3 years ago

8 0

Answer:

50 cm

Explanation:

igomit [66]3 years ago

4 0

Answer:

Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.

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A 15.5 kg block is pulled by two forces. The first is 11.8N at a 53.7 angle and the second is 22.9 N at a -15.8 angle. What is t

anygoal [31]

Answer:

α = 6.43° (Angle respect to the horizontal axis-x)

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces must be equal to the product of mass by acceleration.

Since we have two forces and we know their magnitudes and their directions, we can sum these into their X & y components, in this way we can find the resulting force and apply it in Newton's second law.

F1x = 11.8*cos(53.7) = 6.98 [N]

F1y = 11.8*sin(53.7) = 9.5 [N]

Now with the second force:

F2x = 22.9*cos (15.8) = 22.03 [N]

F2y = - 22.9*sin (15.8) = - 6.23 [N]

Now we sum the forces in the x and y axes:

Fx = F1x + F2x = 6.98 + 22.03 = 29.01 [N]

Fy = F1y + F2y = 9.5 - 6.23 = 3.27 [N]

Now using the Pythagorean theorem we can find the resulting force.

F = √(Fx² + Fy²)

F = √ (29.01)² + (3.27)²

F = 29.19 [N]

Using Newton's second law, we have:

F = m*a

where:

F = force = 29.19 [N]

m = mass = 15.5 [kg]

a = acceleration [m/s²]

a = 29.19/15.5

a = 1.88 [m/s²]

The direction of the acceleration is the same direction of the force, therefore we need to find the angle.

tan(α) = Fy/Fx

tan(α) = 3.27/29.01

α = tan⁻¹(0.1127)

α = 6.43° (Angle respect to the horizontal axis-x)

3 0

3 years ago

A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant

OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

$a_t = ar$

where;

a is the angular acceleration and

r is the radius of the circular path

$a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2$

Determine time of the rotation;

$\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\$

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

$a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2$

The magnitude of total linear acceleration is given by;

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2$

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0

3 years ago

Which of the outer planets is MOST like Earth? How?

Over [174]

Answer:

It is not impossible to answer the question, my apologies for “discrediting” the answer it would have to be Venus, the outer planet most similar to earth.

Explanation:

Venus is very similar to earth, as many studies, research, factual intellectual based concepts - the size, shape, surface gravity, average density, mass, average days of 24 hours, season changes similar to earth, and tilted axis. As well as similar climate make Venus the most accurate answer.

Hope this helps! Thank you for your time.

5 0

4 years ago

Read 2 more answers

As you drive in your car at 22.5 mi/h, you see a child’s ball roll into the street ahead of you. You hit the brakes and stop as

evablogger [386]

Answer:

Explanation:

Speed of car =22.5miles/hr

U=22.5miles/hour

Applied brake and come to rest

Final velocity, =0

t, =2sec

Given that,

Speed=distance /time

Then,

Distance, =speed, ×time

Converting mile/hour to m/s

Given that

Use: 1 mile= 1600 m, 1 h= 3600s

22.5miles/hour × 1600m/mile × 1hour/3600s

Therefore, 22.5mile/hour=10m/s

Using speed =10m/s

Distance =speed ×time

Distance=10×2

Distance, =20m

The distance travelled before coming to rest is 20m.

5 0

3 years ago

The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,

Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

$Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N$

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

$a=Fr/m$

$a=-4291.875/1750\\a=-2.4525$

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

$v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m$

3 0

3 years ago