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kirill [66]
4 years ago
12

The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after t

he driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? a. During each of the three different stages of its motion, the car is moving with constant acceleration. b. During each of the three different stages of its motion, the car is moving with constant velocity. c.The highway is straight (i.e., there are no curves). d. The highway is level (i.e., there are no hills or valleys).Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD.
Physics
1 answer:
jolli1 [7]4 years ago
8 0

Answer:

B, C, D are valid assumptions

Explanation:

In each stage the acceleration of the car varies because the velocity of car varies in each stage. At first, the acceleration of the car is moving slowly.

Then, driver realizes that he is late so the acceleration of the car increases he hits the gas.

Finally, the acceleration decreases when he sees the police car.

In the first state when driver do not know, that he is late, he will drive with a constant velocity as any one does.

When the driver hits the gas and he does not know about the police vehicle, even then he may drive with constant velocity or he may accelerate the car due to fear of being caught.

In the last part of motion when driver see the police car, even than he may accelerate the car, but acceleration will be constant throughout the motion is not possible, or even than he may continue with constant velocity.

Hence, there are only three assumptions are valid.

B. During each of the three different stages of its motion, the car is moving with constant velocity.

C . The highway is straight.

D. The highway is level.

Note that the rate of change of speed is equal to the acceleration. The acceleration is constant if change in velocity of particle is equal in equal interval of time.

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A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0
4 years ago
If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu
ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , n=1.34 \times 10^{20} .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in  one electron is :

e^-=-1.6\times 10^{-19}\ coulombs

So , charge on n electron is :

C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
Leno4ka [110]

Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

7 0
3 years ago
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Which sequence shows the chain of energy transfers that create surface currents on the ocean?
Papessa [141]

Answer:

The correct answer is A. The sun is the energy source of the surface currents in the ocean

6 0
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