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Dimas [21]
3 years ago
10

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

the dimension of time. Determine the dimen- sions of the constants A and B. (b) Determine the dimen- sions of the derivative dx/dt 5 3At2 1 B.
Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) \frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

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A dog pushes against the front door for 3 seconds with a force of 88 N. What
kirill [66]

264Ns

Explanation:

Given parameters:

Time of the push = 3s

Force of push = 88N

Unknown:

Impulse = ?

Solution:

Impulse is defined as the change in momentum of a body when force acts on it.

  Impulse = Force x time

Inputting the parameters:

    Impulse = 88 x 3 = 264Ns

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

5 0
2 years ago
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m
umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s

v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads

Now, we just use the formula:

w_f^2-w_o^2=2\alpha*x

\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2

6 0
2 years ago
Read 2 more answers
From her bedroom window a girl drops a water-filled balloon to the ground, 4.60 m below. If the balloon is released from rest, h
anygoal [31]
Known variables
d=4.6m
initial velocity=0m/s
downward acceleration=-9.8m/s2

d=1/2gt2
4.6=1/2 -9.8 t2
t=0.93s

8 0
3 years ago
read the top corner of picture for directions AWNSER ALL CORECTLY I WILL MAR BRAINLIEST FOR FIRST ANSWER
omeli [17]

Answer:

question

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7 0
3 years ago
2. An archer shoots an arrow at 83.0 m/s at a 62.0 degree angle. If the ground is flat, how much time is the arrow in the air?
UkoKoshka [18]

Answer:

<em>t=14.96 sec</em>

Explanation:

<u>Diagonal Launch </u>

It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.

The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is :

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the initial speed, \theta is the angle, t is the time and g is the acceleration of gravity .

In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus y_o=0, and:

\displaystyle y=v_osin\theta \ t-\frac{gt^2}{2}

The value of y is zero twice: when t=0 (at launching time) and in t=t_f when it goes back to the ground. We need to find that time t_f by making y=0

\displaystyle 0=v_osin\theta\ t_f-\frac{gt_f^2}{2}

Dividing by t_f

\displaystyle v_osin\theta=\frac{gt_f}{2}

Then we find the total flight time as

\displaystyle t_f=\frac{2v_osin\theta}{g}

\displaystyle t_f=\frac{2(83)sin\ 62^o}{9.8}

\displaystyle t_f=14.96\ sec

5 0
3 years ago
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