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jasenka [17]
3 years ago
5

Calculate the acceleration of a car that is maintaining a constant velocity of 1m/s

Physics
2 answers:
lora16 [44]3 years ago
5 0
"Constant velocity" is another way of saying "zero acceleration".
Papessa [141]3 years ago
3 0

Answer:

Acceleration, a =0

Explanation:

If the velocity is constant, it means the acceleration of the object is equal to zero. Mathematically, the acceleration of an object is given by :

a=\dfrac{dv}{dt}

It is also defined as the change in speed per unit time. Its SI unit is m/s^2. It is a vector quantity.

So, it is concluded that if a car is maintaining a constant velocity of 1 m/s, its acceleration is equal to zero. Hence, this is the required solution.

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In the two-slit experiment, monochromatic light of wavelength 600 nm passes through a 19) pair of slits separated by 2.20 x 10-5
kumpel [21]

Explanation:

It is given that,

Wavelength of monochromatic light, \lambda=600\ nm=6\times 10^{-7}\ m

Slits separation, d=2.2\times 10^{-5}\ m

(a) We need to find the angle corresponding to the first bright fringe. For bright fringe the equation is given as :

d\ sin\theta=n\lambda, n = 1

\theta=sin^{-1}(\dfrac{\lambda}{d})

\theta=sin^{-1}(\dfrac{6\times 10^{-7}}{2.2\times 10^{-5}})

\theta=1.56^{\circ}

(b) We need to find the angle corresponding to the second dark fringe, n = 1

So, d\ sin\theta=(n+\dfrac{1}{2})\lambda

sin\theta=\dfrac{3\lambda}{2d}

\theta=sin^{-1}(\dfrac{3\lambda}{2d})

\theta=sin^{-1}(\dfrac{3\times 6\times 10^{-7}}{2\times 2.2\times 10^{-5}})

\theta=2.34^{\circ}

Hence, this is the required solution.

4 0
4 years ago
The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.
melamori03 [73]
If you subscribe I’ll answer QF Aotrx
8 0
4 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
4 years ago
a spring is hanging down from the ceiling, and an object of mass m is attached to the free end. the object is pulled down, there
Maurinko [17]

The dimension of K is M/ T^2

according to the question T=2π square root ofm/k here 2 pi is constant so  

T= root of m /k and root of k = root of m/ T now by squaring on both the sides we get the answer k= M/ T^2

complete question :

A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T=√2πm/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?

To learn more about dimension:

brainly.com/question/13314350

#SPJ4

7 0
2 years ago
. A new game has been invented where the player has to combine a dive from a board with shooting of a basketball through a hoop
skad [1K]

Explanation:

Not being able to see the picture attached and supposing the ground level is considered to be the water level so to be able to find out the velocity you need to score into the hoop you need to use the combined force ( resulting force) to throw the ball at 6.50 m distance, 1 m below to where you are standing

5 0
3 years ago
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