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AysviL [449]
3 years ago
14

Holden is trying to turn man the velocity of his car, he went 20 meters east, turned around and left 40 meters less, he turned t

he car one more time, and when 35 meters east, his car was 15 metres from the starting line. This took 5 seconds, what is the car
velocity
Physics
1 answer:
Grace [21]3 years ago
4 0

Answer:

3 m/s east

Explanation:

The velocity of the car is given by:

v=\frac{d}{t}

where

d is the displacement

t is the time taken

The displacement of the car does not depend on the path of the car, but just on it the difference between its final position and its starting position, so it is equal to 15 metres east. The time taken is 5 seconds, therefore the velocity is equal to

v=\frac{15 m}{5 s}=3 m/s

and the direction is east.

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You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
2 years ago
Thermal energy depends on an object’s
oee [108]

The correct answer is D. I alread took this test.

4 0
3 years ago
Read 2 more answers
a watermelon is balanced by a dog, a pumpkin, a flowerpot, and a baseball as shown below. Is the weight of the watermelon equal
alukav5142 [94]

Answer:

pumpkin

Explanation:

watermelon and pumpkins are close to shape and size

6 0
2 years ago
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Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/s^{2}, t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

  = 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

 = 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + \frac{1}{2}at^{2}

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = \frac{1}{2}at^{2}

  = \frac{1}{2} x 14 x (510)^{2}

 = 7 x 260100

 = 1820700 m

The distance that the shuttle has traveled during the given time is  1820.7 Km.

5 0
2 years ago
In a laboratory, you determine that the density of a certain solid is 5.23× 10 −6 kg/m m 3 . convert this density into kilograms
tamaranim1 [39]

Density is given as

\rho = 5.23 * 10^{-6} kg/mm^3

now we have to convert this density into kg/m^3

now we have

1 m = 1000 mm

\rho = 5.23 * 10^{-6} \frac{kg}{(1*10^-3 m)^3}

\rho = 5.23 * 10^{-6} \frac{1*10^9 kg}{m^3}

\rho = 5.23 * 10^3 kg/m^3

\rho = 5230 kg/m^3

8 0
3 years ago
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