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3 years ago

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when a man throws a stone in the direction of maximum range the stone rises to a height of 20m . what is the range and how long

is the stone in air

1 answer:

jeyben [28]3 years ago

3 0

Answer:

50m

Explanation:im just smart thank me later

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so i did this thing and idk how it happened but basically i put foil inside the tip of my computer charger while it was charging

just olya [345]

Answer:

It was most likely static electricity

Explanation:

Static electricity comes down to the interactive force between electrical charges.

So because your computer charger has electricity the foil is a conductor so the sparks were most likely static electricity

3 0

3 years ago

A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the

Rudik [331]

Answer:

Q must be placed at 0.53 L

Explanation:

Given data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at distance x cm from q1

The force by charge q_1 due to q is

$F1 = \frac{k q q_1}{x^2}$

$F1 = \frac{k q ( 4.0 μC )}{ x^2}$ ----1

The force by charge q_2 due to q is

$F2 = \frac{k q q_2}{(L-x)^2}$

$F2 = \frac{kq (3.0 μC)}{(L-x)^2}$ --2

we know that net electric force is equal to zero

F_1 = F_2

$\frac{k q ( 4.0 μC )}{x^2} =\frac{k q ( 3.0 μC )}{(l-x)^2}$

$\frac{4}{3}*(L-x)^2 = x^2$

$x = \sqrt{\frac{4}{3}*(L - x)$

$L-x = \frac{x}{1.15}$

$L = x + \frac{x}{1.15} = 1.86 x$

x = 0.53 L

Q must be placed at 0.53 L

3 0

2 years ago

Which two surfaces have the lowest coefficient of friction

jok3333 [9.3K]

Answer:

Explanation:

As an example, ice on steel has a low coefficient of friction – the two materials slide past each other easily – while rubber on pavement has a high coefficient of friction – the materials do not slide past each other easily. The coefficients of friction ranges from near 0 to greater than 1.

5 0

2 years ago

A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c

Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; $m_{total$ v

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

3 0

2 years ago

A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.

dsp73

Answer:

<h3>0.445</h3>

Explanation:

In friction, the coefficient of friction formula is expressed as;

$\mu = \frac{F_f}{R}$

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

$\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta$

Given

θ = 24°

$\mu = tan 24^0\\\mu = 0.445\\$

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0

3 years ago