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solmaris [256]
4 years ago
9

A car traveling east in a straight line on a highway decreases its speed from 30. meters per second to 23. meters per second in

2.0 seconds. What is the magnitude of the car's average acceleration to the nearest tenths place, during this time interval?
Physics
1 answer:
Lera25 [3.4K]4 years ago
7 0

Answer:

-3.5m/s²

Explanation:

Given

Initial Velocity, u = 30m/s

Final Velocity, v = 23m/s

Time, t = 2.0s

Required

Determine the magnitude of acceleration

This is calculated using first equation of motion.

v = u + at

Substitute values for v, u and t

23 = 30 + a * 2.0

23 = 30 + 2.0a

23 - 30 = 2.0a

-7 =2.0a

Solve for a

a = -7 ÷ 2.0

a = -3.5m/s²

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A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co
Rus_ich [418]

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N

So, the required force is 319.2 N.

3 0
3 years ago
Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir
strojnjashka [21]

Answer:

1.475\times 10^{-13}\ C/m^3

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area

h = Altitude = 600 m

Electric flux through the top would be

-110A (negative as the electric field is going into the volume)

At the bottom

120A

Total flux through the volume

\phi=120-110\\\Rightarrow \phi=10A

Electric flux is given by

\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0

Charge per volume is given by

\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3

The volume charge density is 1.475\times 10^{-13}\ C/m^3

7 0
3 years ago
If an organism didn't have DNA, it would be missing which sign of life ? a, b , c
sergey [27]

Answer:

A

Explanation:

4 0
3 years ago
Read 2 more answers
An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the
Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image​. Let v be the image distance. Using lens formula to find it :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put all the values,

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm

So, the image distance from the lens is 24 cm.

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2

The negative sign of magnification shows that the formed image is inverted.

7 0
3 years ago
Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
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