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Talja [164]
3 years ago
7

The energy lost to friction is destroyed.TrueFalse

Physics
1 answer:
leva [86]3 years ago
5 0
False!!!!!!!!!!!!!!!!!!
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A 19.0-kg cart is moving with a velocity of 7.20 m/s down a level hallway. A constant force of -13.0 N acts on the cart and its
gulaghasi [49]

Answer:

a. -369.36J

b. -123.9J

c. 9.52m

Explanation:

From the expression for kinetic energy

K. E=1/2mv^2

Since the mass is constant, but the velocity changes. Hence the change in kinetic energy is

K.E=1/2*19(3.6²-7.2²)

K.E= -369.36J

b. to determine the workdone by the force,we determine the distance moved.

But the acceleration is from

F=ma ,

a=f/m

a=-13/19

0.68m/s²

the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

Hence the work done is

W=force * distance

W=-13*9.52

W=-123.9J

d. the distance moved is

s=v²/2a

s=3.6²/2*0.68

s=9.52m

4 0
4 years ago
Read 2 more answers
An example of a substance with a definite volume and a definite shape is
kolezko [41]
A. a steel beam. All of the rest of them change shape or volume whereas a steel beam with never change in volume or shape.
7 0
4 years ago
So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

5 0
3 years ago
Calculate the kinetic energy of a 1 kg object moving at 4.5 ms. Round to the first decimal.
harina [27]

Answer:

91.21

Explanation:

Kinetic energy =1/2 mv2 substitute the values I. E 1/2 *1 *4.5^2= 91.21

8 0
3 years ago
A circuit consists of a 6 ohm resistor, a 0.2 farad capacitor, and an AC voltage source supplying V(t) = 120 cos(20 t) volts. Wr
Zinaida [17]

Answer:

q = 24 cos (20t) (1- e (-t / 1.2))

Explanation:

To work this circuit we use the mesh equation (Kirchoff) that states that the voltage in a closed circuit is zero

           Vg + Vr + Vc = 0

Where Vg is the generator voltage, Vr the resistance voltage and Vc the capacitor voltage

         Vg = 120 cos 20t  = V

         Vr = i R

         Vc = q / C

We see that in one term we have the current (i) and in another the load (q), but there is a relationship between the two

         i = dq / dt

Let's replace in the initial equation

       V + R dq / dt + q / C = 0

Reorder the terms

      Rdq / dt + q / C - V = 0

      dq / dt + q / rC - V / R = 0

      dq / dt = V / r -q / RC

       dq / (V / R -q / RC) = dt

we integrate

     ∫I dq / (V / R - 1 / RC q) = ∫ dt

We change variables

      u = (V / R - 1 / RC q)

      du = -1 / RC dq

     ∫ dq / (V / R - 1 / RC q) = -RC ∫ du / u

     -RC ln u = -RC ln (V / R - 1 / RC q)

We evaluate between the limits of integration, the lower t = 0 q (0) = 0

     -RC [ln (V / R - 1 / RC q) - ln (V / R)] = t

     [ln (V / R - 1 / RC q) - ln (V / R)] = -t / RC

     Ln (V / R - 1 / RC q) / (V / R)] = -t / RC

     Ln (1 - 1 / VC q)) = (-t / RC)

     Ln (VC- q) = ln (VC) (-t / RC)

     VC-q = VC e (-t / RC)

     q = VC - Vc e (-t / RC)

     q = VC (1- e (-t / RC)

We substitute the values ​​they give us

      q = (120 cos (20t) 0.2) (1- e (-t / 6 0.2))

      q = 24 cos (20t) (1- e (-t / 1.2))

7 0
4 years ago
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