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3 years ago

A race car circles 10 times around a circle with a diameter of 7 km track in 50 min. Using SI

Physics

1 answer:

vodka [1.7K]3 years ago

4 0

Average Speed = 23.33 m/s

<u>Explanation:</u>

Given:

Diameter = 7 km

So, to find the distance, we need to multiply it by 10 as the race car circles 10 times round this 7 km track.

Distance = 7 × 10 = 70 Km or 70,000 m

Time = 50 mins

= 50 × 60 = 3000 s

Now, in order to find the average speed for 10 cycles, we use the formula:

$Speed = \frac {Distance} {Time}$

So by substituting the above values in the given formula, we get:

$Speed = \frac {70,000} {3000}$

Speed = 23.33 m/s

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Richard is driving home to visit his parents. 135{\rm mi} of the trip are on the interstate highway where the speed limit is 65{

nydimaria [60]

Answer:

time spent = 0.2276

Explanation:

given data

distance = 135 mi

usual speed = 65 mph

today speed = 73 mph

solution

we get here time that is express as

time = $\frac{distance}{speed}$ ...................1

usual time = $\frac{135}{65}$ = 2.0769 h

today time = $\frac{135}{73}$ = 1.8493 h

so we get here time spent as

time spent = 2.0769 h - 1.8493 h

time spent = 0.2276

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3 years ago

Why do you like the full moon ?

mr_godi [17]

Answer:

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4 0

3 years ago

Read 2 more answers

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

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Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

3 years ago

Two ions with masses of 4.39×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is

Ilia_Sergeevich [38]

Answer:

7.2 cm

Explanation:

magnetic field, B = 0.301 T

speed, v = 7.92 x 10^5 m/s

mass, m = 4.39 x 10^-27 kg

q = 1.6 x 10^-19 C

The radius of singly changed ion is given by

$r = \frac{mv}{Bq}$

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

$r = \frac{4.39\times 10^{-27}\times 7.92 \times 10^{5}}{0.301\times 1.6\times 10^{-19}}$

r = 0.072 m

r = 7.2 cm

5 0

3 years ago