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Ivan
4 years ago
8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?
Physics
1 answer:
pychu [463]4 years ago
5 0
(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²
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An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha
hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

F_D =F_g

By definition we know that the Drag force is defined as

F_D= \frac{1}{2} C_d \rho A V^2

Where,

C_d = Drag coefficient

\rho =Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

F_g = (m_1 +m_2) g

Where,

m_1 =Mass of sphere

m_2 =Mass of unknown object

Equating the two equations we have to

(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2

Re-arrange for m_2,

m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1

Our values are given by,

C_d = 0.5

\rho = 1.22Kg/m^3

V = 66.7m/s

m_1 = 3Kg

d= 32.7*10^{-2}m

r = 16.35*10^{-2}m

Replacing in the equation we have,

m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3

m_2 = 8.62Kg

<em>Therefore the mass of unknown object is 8.62Kg</em>

4 0
3 years ago
A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of 2 newtons. What is the accel
notka56 [123]

Explanation:

F = ma

2 N = (10 kg) a

a = 0.2 m/s²

6 0
3 years ago
2. Another car weighs 2000kg, you can push it .05 m/s?, how much force are you
Vesna [10]

Answer:

\boxed {\boxed {\sf 100 \ Newtons}}

Explanation:

We are asked to calculate the force you are applying to a car. According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, we can use the following formula to calculate force.

F= m \times a

The mass of the car is 2000 kilograms and the acceleration is 0.5 meters per second squared.

  • m= 2000 kg
  • a= 0.05 m/s²

Substitute the values into the formula.

F= 2000 \ kg \times 0.05 \ m/s^2

Multiply.

F= 100 \ kg *m/s^2

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 100 kilogram meters per second square is equal to 100 Newtons.

F= 100 \ N

You apply <u>100 Newtons</u> of force to the car.

7 0
3 years ago
A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and t
Zina [86]

Answer:

2000 W

Explanation:

First of all, we need to find the output voltage in the transformer, by using the transformer equation:

\frac{V_1}{N_1}=\frac{V_2}{N_2}

where here we have

V1 = 200 V is the voltage in the primary coil

V2 is the voltage in the secondary coil

N1 = 250 is the number of turns in the primary coil

N2 = 500 is the number of turns in the secondary coil

Solving for V2,

V_2 = N_2 \frac{V_1}{N_1}=(500) \frac{200 V}{250}=400 V

Now we can find the power output, which is given by

P = VI

where

V = 400 V is the output voltage

I = 5 A is the output current

Substituting,

P = (400 V)(5 A) = 2,000 W

3 0
4 years ago
What is the length of the blues progression in this excerpt?
Gre4nikov [31]
<span>In blues musicians also often refer to chord progressions using Roman numerals, as this facilitates transposing a song to a new key. For example, rock and blues musicians often think of the 12 bar blues as consisting of I, IV and V chords. Thus a simple version of the 12 bar blues might be expressed as I/I/I/I IV/IV/I/I V/IV/I/I. By thinking of this blues progression in Roman numerals, a backup band or rhythm section could be instructed by a band leader to do the chord progression in any key. For example if the band leader asked the band to play this chord progression in the key of C Major, the chords would be C/C/C/C F/F/C/C G/F/C/C.</span>
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3 years ago
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