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2 years ago

A ball dropping from 70 cm took 0.6 seconds to travel a horizontal distance of 34 cm.

Physics

1 answer:

Alenkasestr [34]2 years ago

4 0

Answer:

vₓ = 0.566 m / s, W_total = 9.1 J

Explanation:

This exercise is a parabolic type movement, for the x axis where there is no acceleration

x = v t

vₓ = x / t

vₓ = 0.34 / 0.6

vₓ = 0.566 m / s

the work done is

X axis

In this axis there is no acceleration, therefore the sum of the forces is zero and since the work is the force times the distance, we conclude that the lock in this axis is zero.

W₁ = 0

Y axis

in this axis the force that exists is the force of gravity, that is, the weight of the body

W₂ = Fy y

W₂ = mg and

W₂ = m 9.8 0.70

W₂ = m 9.1

the work is a scalar for which we have to add the quantities obtained

W_total = W₁ + W₂

W_total = 0 + 9.1 m

W_total = 9.1 m

In order to complete the calculation, the mass of the body is needed if we assume that the mass is m = 1

W_total = 9.1 J

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What is the net force in the following diagram?

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2 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 51.3°. The veloci

abruzzese [7]

Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.

The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
163° - 51.3° = 111.7°

Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
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3 years ago

Read 2 more answers

A current exists whenever electric charges move. If ΔQ is the net charge that passes through a surface during a time period Δt,

jeka57 [31]

Answer:

It represents the change in charge Q from time t = a to t = b

Explanation:

As given in the question the current is defined as the derivative of charge.

I(t) = dQ(t)/dt ..... (i)

But if we take the inegral of the equation (i) for the time interval from t=a to

t =b we get

Q =∫_a^b▒〖I(t) 〗 dt

which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.

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3 years ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago

A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =

stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

$d_{13} =\sqrt{24.68}$ * 10⁻²m = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

F₂₃x = F₂₃ = +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

$F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2} }$

$F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2} }$ *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0

3 years ago