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ivanzaharov [21]
3 years ago
11

A dragster is traveling at a constant rate of 45 m/s. How fast is it going when it has traveled 325 m?

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0

The speed of the dragster after travelling 325 m is still 45 m/s

Explanation:

In this problem, we are told that the dragster is travelling at constant rate, therefore its speed is constant and it is

v = 45 m/s

Therefore, its speed will remain constant during the entire motion: therefore, after having travelled for 325 m, its speed will still be 45 m/s.

We can also calculate the time taken for the dragster to cover the 325 m. In fact, the speed in a uniform motion is given by

v=\frac{d}{t}

where:

v is the speed

d is the distance covered

t is the time elapsed

In this case, we have:

v = 45 m/s

d = 325 m

Solving for t, we find:

t=\frac{d}{v}=\frac{325}{45}=7.2 s

Learn more about speed:

brainly.com/question/8893949

#LearnwithBrainly

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41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h
romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0
3 years ago
An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t
Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

At one point

Wide= 16.0 m

Deep = 3.8 m

Water flow = 2.8 cm/s

At a second point downstream

Width of canal = 16.5 m

Water flow = 11.0 cm/s

We need to calculate the depth

Using Bernoulli theorem

A_{1}V_{1}=A_{2}V_{2}

Put the value into the formula

16.0\times3.8\times2.8=16.5\times x\times 11.0

x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}

x=0.938\ m

Hence,  The depth of the water at this point is 0.938 m.

7 0
3 years ago
Drag each tile to the correct location.
scoray [572]

Answer:

Relative: meredith's birth, the planting of jasmine's tree, and the extinction of the stegosaurus

Absolute: harrison's birth, the building of jasmine house, and the extinction of the triceratops

Explanation: plato

7 0
2 years ago
A heat engine with 0.500 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 700 K . The gas goes through the fo
Zarrin [17]

Answer:

2234.63

Explanation:

Work done per cycle by the engine is calculated as;

W=\int p \ dv

#Since volume doesn't change in the isochoric steps, there is no work done, hence:

W_2=W_3=0

#For an isothermal change of state(ideal gas):

pV=nRT\\\\W=P_iV_ i\int \frac{1}{V} \ dV\\\\=P_iV_ iIn(V_f/V_i)

#for the expansion process:

W_i=0.5mol\times8.3145J/molK\times 700K \ In(6000cm^3/1000cm^3)\\\\W_i=5214.15J

W_4=0.5mol\times8.3145J/molK\times 400K \ In(6000cm^3/1000cm^3)\\\\W_4=-2979.52

W=W_1+W_2+W_3+W_4\\\\=5214.15J-2979.52\\\\=2234.63J

Hence, the engine does 2234.63J per second.

8 0
3 years ago
When the power fails completely, the battery in a(n) ___ contains enough power to run a computer system for a given amount of ti
e-lub [12.9K]

i think the answer is neutron

:)

4 0
2 years ago
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