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elena-14-01-66 [18.8K]
2 years ago
12

Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f

inal speed when sliding off? Assume that there is no friction acting.
Physics
1 answer:
Lunna [17]2 years ago
8 0

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = \frac{1}{2}mv^{2}

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

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

final velocity of the child , v = \sqrt{2gh}

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

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Physical forms are: gas,liquid,and solid
8 0
2 years ago
A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t
choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25
8 0
2 years ago
Read 2 more answers
A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa
kipiarov [429]

Answer:

2.45 m

Explanation:

First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:

h=\frac{1}{2}gt^2

where

h = 1.19 m is the vertical distance covered by the book

g = 9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s

Now we can find the horizontal distance covered by the book, which is given by

d=v_x t

where

v_x = 5.0 m/s is the horizontal velocity

t = 0.49 s is the time of flight

Substituting,

d=(5.0)(0.49)=2.45 m

So the book lands 2.45 m away.

8 0
2 years ago
A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca
Hatshy [7]

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

u=\frac {v^{2}}{2gd}

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367

Therefore, the coefficient of kinetic friction is approximately 0.0367

7 0
2 years ago
A coaxial cable consists of a thin insulated straight wire carrying a current of 2.00 A surrounded by a cylindrical conductor ca
ella [17]

Answer:

B = 15μT

Explanation:

In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere's law, which is given by:

B=\frac{\mu_oI}{2\pi r}       (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current

r: distance from the wire to the point in which B is calculated

In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:

B_T=B_1-B_2=\frac{\mu_o I_1}{2\pi r}-\frac{\mu_o I_2}{2\pi r}\\\\B_T=\frac{\mu_o}{2\pi r}(I_1-I_2)  (2)

I1: current of the central wire = 2.00A

I2: current of the cylindrical conductor = 3.50A

r: distance = 2.00 cm = 0.02 m

You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.

|B|=|\frac{4\pi*10^{-7}T/A}{2\pi (0.02m)}(2.00A-3.50A)|=1.5*10^{-5}T\\\\|B|=15*10^{-6}T=15\mu T

The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT

3 0
2 years ago
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