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docker41 [41]
3 years ago
9

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the displacement of the object?

Physics
2 answers:
Sloan [31]3 years ago
5 0

In a vleocity Vs time graph, the change in position of an object is given by

x_{1}-x_{2} =\int\limits^T_t {v} \, dt

where the integral can be taken from the graph as

\int\limits^T_t {v} \, dt=area between velocity curve and time axis from t to T


forsale [732]3 years ago
4 0
The area of the rectangle plus the area of the triangle under the line .-.
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7. If the magnitude of the gravitational force of
JulijaS [17]

Newton's Third Law states that for every action there is an opposite and equal reaction:

If the gravitational force of the Earth on the Moon is F then the gravitational force of the Moon on the Earth is also F

3 0
2 years ago
If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1,000 km
astra-53 [7]

Answer:

change 1000km into Metre

1000km=1000000 M

speed= distance/time

time=distance/speed

time=1000000/299792458

time= 0.0033second

6 0
2 years ago
A student is wearing a pair of sunglasses designed to reduce the glare from reflected surfaces. When the student tilts her head
agasfer [191]
The answer is polarization. It is just like when you are on a boat and use polarized glasses so the water wont reflect into your eyes it also intensifies the color of the water so you can see better into the water.
example: you cant see any fish because glare from the sun, put on polarized glasses and now you can see fish up to 30 feet deep.
6 0
3 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the
mamaluj [8]

Answer:

<h2> 1.643*10⁻⁴cm</h2>

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = \frac{\delta m \lambda d}{a} where;

\delta m is the first two diffraction minima = 1

\lambda is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given \lambda = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

3 0
3 years ago
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