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2 years ago

A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?

2 answers:

5 0

Simply subtract the two velocities and divide by 8.1,

$\frac{0 - 35}{8.1} = - 4.32$

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey

Ne4ueva [31]2 years ago

4 0

Answer:

The acceleration of the car is -4.32 m/s^2.

Explanation:

Given that,

Speed of car u= 35 m/s

Time t = 8.1 s

We need to calculate the acceleration

using equation of motion

$v = u+at$

Where, v = final velocity

u = initial velocity

t = time

a = acceleration

$0=35+a\times8.1$

$a = -\dfrac{35}{8.1}$

$a = -4.32\ m/s^2$

Hence, The acceleration of the car is -4.32 m/s^2.

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A professor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges.

taurus [48]

In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it. All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges. That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

4.9 N-m = (force) x (0.89 m)

Divide each side by 0.89m: Force = (4.9 N-m) / (0.89 m)

= 5.506 N .

7 0

2 years ago

You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg

Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball $m=0.0570\ kg$

ball leaves the hand with $u=15\ m/s$

maximum height reached by ball $h=8\ m$

Initial Mechanical energy when ball just leaves the hand

$M.E._1=(P.E.+K.E.)_1$

$M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1$

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

$M.E._1=\frac{1}{2}\times m\times (15)^2$

$M.E._1=6.41\ J$

Mechanical Energy at highest point

$(M.E.)_2=(P.E.+K.E.)_2$

at highest Point velocity is zero

$(M.E.)_2=mgh_2+0$

$(M.E.)_2=0.0570\times 9.8\times 8$

$(M.E.)_2=4.46\ J$

Decrease in Mechanical energy

$(M.E.)_1-(M.E.)_2=6.41-4.46$

$(M.E.)_1-(M.E.)_2=1.95\ J$

3 0

2 years ago

A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass

Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

$L=2\pi R$

where R is the radius of the wheel.

The period of revolution is:

$T=120 s$

Therefore, the tangential speed of the book is:

$v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R$

8 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

Calculate the longest wavelength visible to the human eye. express the wavelength in nanometers to three significant figures.

slega [8]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

The human eye contains a molecule called 11-cis-retinal that changes conformation when struck with light of sufficient energy. The change in conformation triggers a series of events that results in an electrical signal being sent to the brain. The minimum energy required to change the conformation of 11-cis-retinal within the eye is about 164 kJ/mole. Calculate the longest wavelength visible to the human eye.

Solution:

Energy (E) = 164 kJ/mole

E = 164 kJ/mole = 164 kJ /6.023 x 10²³

= 2.72 x 10⁻²² kJ = 2.72 x 10⁻¹⁹J

Planck's constant = 6.6 x 10⁻³⁴ J s,

Speed of light = 3.00 x 10⁸ m/s

Let the required wavelength be λ.

Formula Used: E = hc / λ

or, λ = hc / E

or, λ = (6.6 x 10⁻³⁴ J s)× (3.00 x 10⁸ m/s) / (2.72 x 10⁻¹⁹J)

or, λ = 7.28 x 10⁻⁷ m

or, λ = (7.28 x 10⁻⁷ m) ×( 1.0 x 10⁹ nm / 1.0 m)

or, λ = (7.28 x 10² nm)

or, λ = 728 nm

Hence, the required wavelength will be 728 nm.

6 0

2 years ago