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sergejj [24]
3 years ago
6

A car is driving on the road. The car starts to decelerate at -3 m/s2. It stops moving after 125

Physics
1 answer:
galina1969 [7]3 years ago
8 0

Answer:

27.39 m/s

Explanation:

From the question given above, the following data were obtained:

Deceleration (a) = –3 m/s²

Distance (s) = 125 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

The initial velocity (u) of the car can be obtained as follow:

v² = u² + 2as

0² = u² + (2 × –3 × 125)

0 = u² + (– 750)

0 = u² – 750

Collect like terms

0 + 750 = u²

750 = u²

Take the square root of both side

u = √750

u = 27.39 m/s

Thus, the initial velocity of the car was 27.39 m/s

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

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6 0
2 years ago
IN THIS FORMULA FOR WATER WHAT DOES THE SUBSCRIPT 2 INDICATE
masha68 [24]
The chemical formular for water is H2O.
The H aspect of the formula stands for hydrogen gas and the subscript 2 which is attached to the H symbol signifies that two atoms of hydrogen are joined together, that is two atom of hydrogen are present.
The chemical formula of water indicates that, two atom of hydrogen react with one atom of oxygen to form one molecule of water.
In chemical formulae, subscripts are normally used to indicate the number of atoms that are present in a molecule.
7 0
4 years ago
Add these measurements, using significant digit rules:<br><br> 1.0090 cm + 0.02 cm = cm
marin [14]

Answer:

1.029

Explanation:

1.0090 can also be looked at as "1.009"

0.02 can also be looked at as "0.020"

I think of it as 20+9 which is 29. There for your answer should be 1.029. There are no measurement rules applying to this equation since they are both in centimeters. So you don't have to convert anything.

7 0
3 years ago
Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th
Alik [6]

Answer:

The wavelength of the light is 7200\ \AA.

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes \beta= 0.3\ cm

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

\beta=\dfrac{D\lambda}{d}

Put the value into the formula

0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}

\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}

\lambda=7.2\times10^{-7}\ m

\lambda=7200\ \AA

Hence, The wavelength of the light is 7200\ \AA.

8 0
3 years ago
A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance  

= mass * gravity * distance

= 1000 * 9.81 * 10  

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J  

Work done by cable = PE +KE  

108,100 J = KE + 98,100 J  

KE = 10,000 J

= 10kJ

=

7 0
3 years ago
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