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3 years ago

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If the speed of an object is tripled, its kinetic energy will be

2 answers:

Savatey [412]3 years ago

7 0

Answer:

if the speed of the object triples the kinetic energy becomes nine times the initial kinetic energy. If the speed of an object changes by a factor k, the change in the kinetic energy is of a factor k^2.

Explanation:

UkoKoshka [18]3 years ago

5 0

Answer:

it will be tripled bc it is going tripled the speed so the kinetic energy will be to hope this help

Explanation:

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A 720 kg roller-coaster starts off from Location A. Assuming friction does not impede the car's motion, what will be the change

vaieri [72.5K]

We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:

ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)

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3 years ago

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A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters

julia-pushkina [17]

Answer:

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3 years ago

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s.

Lyrx [107]

Answer:

a) 1.22 s

b) 9.089 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-9.81}\\\Rightarrow t=1.22\ s$

Time taken by the ball to reach the highest point is 1.22 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.339\ m$

The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m

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3 years ago

Which example shows potential energy? A. a skydiver falling B. a car racing C. hitting a nail with a hammer D. a wound up watch

tankabanditka [31]

B.a car racing down a hill

6 0

3 years ago

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dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

3 years ago