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grigory [225]
3 years ago
5

Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Dete

rmine how far vertically up the hill the ball reaches when it stops. Show your work.(2 points)
Physics
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

The maximum height is 0.33 m.

Explanation:

initial velocity, u = 8 m/s

final velocity, v = 0 m/s

10% of  kinetic energy is lost in friction.

The kinetic energy used to move up the top,

KE = 10 % of 0.5 mv^2

KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m

Let the maximum height is h.

Use conservation of energy

KE at the bottom = PE at the top

3.2 m = m x 9.8 x h

h = 0.33 m  

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ladessa [460]

Answer:

direct current hope this helped : )

6 0
3 years ago
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Katena32 [7]

Answer:

While slavery was the major issue separating the North and South, it was not slavery itself that sparked the conflict. The South wanted to secede from the Union, and the North refused. While President Abraham Lincoln personally opposed slavery, he recognized that it was legal under the U.S. Constitution at the time. He also recognized that few in the North were ready to go to war to free the slaves. For Lincoln and the northern majority, preservation of the Union was the foremost goal.

4 0
3 years ago
A child on a swing falls a height of 1m. What speed will the child reach?
kirill [66]

Answer:v=4.42\ m/s

Explanation:

Given

Child falls a height of h=1\ m

Applying conservation of energy  we can write

Loss in Potential energy=Gain in kinetic energy

mgh=\frac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 1}

v=\sqrt{19.6}

v=4.42\ m/s

Thus child reaches a velocity of 4.42 m/s

7 0
4 years ago
A toy of mass 0.190-kg is undergoing SHM on the end of a horizontal spring with force constant k = 350 N/m . When the toy is a d
vagabundo [1.1K]

Answer

a)0.0495 J

b)0.01681 m

c)0.7218 m/s

Explanation:

Given

Mass of the.toy M = 0.190 kg

force constant k = 350 N/m

Displacement from equilibrium x = 0.0140 m

Speed v = 0.400 m/s

a)What is the toy's total energy at any point of its motion?

The total energy at any point of it's motion can be calculated by adding together both the potential and kinetic energy of the toy, since it's posses potential energy when at rest and kinetic energy at motion

Total energy E = kinetic energy + potential energy

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.190)(0.4)² + ¹/₂ (350)(0.0140)²

E = 0.0495 J

Hence,the total energy is 0.0495 J

b) the amplitude of the motion can be calculated using below formula

Let amplitude = A

E = ¹/₂KA²

if we make Amplitude A the subject of the formula we have

A=√(2E/k)

But we have calculated our E up there, our K was given in question then if we substitute we have

A= √(2×0.0495)/350

Ans: 0.01681 m

Hence, our Amplitude is 0.01681 m

c) the the toy's maximum speed during its motion can be calculated using the expression below

Let maximum speed = vmax

E = (1/2)M * vmax^2

If we make vmax the subject of the formula we have

vmax =√(2E/m)

vmax= √(2×0.0495)/0.190

vmax=0.7218 m/s

Hence our vmax is 0.7218 m/s

8 0
3 years ago
A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final
77julia77 [94]

Answer:

final velocity = 3504 m/s  

Explanation:

<em>Given data:</em>

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

<em>To find:</em>

Final velocity: ?

<em>Formula:</em>

x = Vx*t + ½*ax*t²  

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

<em>Solution:</em>

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

<em>similarly vertically:</em>

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  

V'y = Vy + ay*t

     = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

V = √(V'x² + V'y²)

   = 3504 m/s  

8 0
3 years ago
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