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DanielleElmas [232]

1 year ago

6

The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee

n t1 = 1.0s and t2 = 4.0s.

1 answer:

aniked [119]1 year ago

5 0

Answer: 15m/s

Explanation: <u>Average</u> <u>Velocity</u> is vector describing the total displacement of an object and the time taken to change its position. It is represented as:

$v=\frac{\Delta x}{\Delta t}$

At t₁ = 1.0s, displacement x₁ is:

$x(1)=25+3(1)^{2}$

x(1) = 28

At t₂ = 4.0s:

$x(4)=25+3(4)^{2}$

x(4) = 73

Then, average speed is

$v=\frac{73-28}{4-1}$

v = 15

The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s

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mrs_skeptik [129]

I think the answer is B but i could be wrong

7 0

2 years ago

Read 2 more answers

The tip of a triangle is held 12.0 cm above the surface of a flat pool of water. A submerged swimmer in the pool sees the tip of

Rufina [12.5K]

Answer: 9cm

Explanation:

Refractive index can also be defined as the ratio of the real depth to the apparent depth.

Given that the

Real depth = 12 m

Refractive index of water = 1.33

Refractive index of air = 1.00

nair/nwater = real depth/apparent depth

Substitute all the parameters into the formula

1.33/1 = 12/ apparent depth

Cross multiply

1.33 Apparent depth = 12

Apparent depth = 12/1.33

Apparent depth = 9.02 cm

Therefore, A submerged swimmer in the pool sees the tip of the triangle at 9cm approximately distance above the water.

8 0

2 years ago

An object is dropped on Earth from a height of 15 m. How long did it take for the object to reach the ground?

DanielleElmas [232]

VA and I will not do this if they don't want it and you are a little

3 0

2 years ago

Assuming the average rate of heat energy flowing outwards from earth to be 0.063 W/m2 calculate the energy available at a hot sp

MaRussiya [10]

Answer:

"13.48 Kwhr" is the right solution.

Explanation:

The given values are:

Average rate of heat energy,

= 0.063 W/m²

Diameter,

= 8m

Efficiency of conversion,

= 50%

Now,

The area of hotspot will be:

⇒ $A=\frac{\pi}{4} d^2$

On substituting the values, we get

⇒ $=\frac{3.14}{4} (8)^2$

⇒ $=0.785\times 64$

⇒ $=50.24 \ m^2$

Total heat generation rate will be:

⇒ $Q=q\times A$

$=0.063\times 50.24$

$=3.16 \ W$

hence,

The electricity generation capacity will be:

⇒ $P=\eta Q t$

On substituting the values, we get

⇒ $=0.5\times 3.16\times 8760$

⇒ $=13840.8 \ Whr$

On converting into Kwhr, we get

⇒ $=13.84 \ Kwhr$

6 0

2 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.360 mm wide. The diffraction pattern is observed

Bumek [7]

Answer:

a) 0.0130 m

b') w' = =6.46*10^{-3] m

Explanation:

given data:

\lambda of light = 633 nm

width of siit a =0.360 mm

distance from screen = 3.75 m

a) the first minima is located at

$sin\theta = \frac{\lambda}{a}$

= $= \frac{633 *10^{-9}}{.360*10^{-3}}$

$\theta = 0.100$

$y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m$

with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m

b)

width of the first bright fringe on either side of the central one = $w' = y_2 -y_1$

calculation for y_2

$sin\theta = 2\frac{\lambda}{a}$

= $= 2*\frac{633 *10^{-9}}{.360*10^{-3}}$

$\theta = 2*0.100 = 0.200$

$y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m$

$w' = 0.0130 -6.54 *10^{-3}$

w' = =6.46*10^{-3] m

6 0

2 years ago