All categories

3 years ago

Why is it difficult to convert miles to yards

2 answers:

6 0

There's really no "why", because it's not difficult at all.

Simply multiply the number of miles by 1,760, and bada boom,
there you have the same distance described in yards.

Lelechka [254]3 years ago

3 0

In today's technological age it would be immensely easy to make such a conversion. There are countless tools on the internet that can do the job for you incredibly quickly and easily. Nevertheless, I'll give you an idea about how it could be done.

$1\quad mile\quad =\quad 1760\quad yards\\ \\ \frac { 1 }{ 1760 } \quad \times \quad 1\quad mile\quad =\quad 1760\quad \times \quad \frac { 1 }{ 1760 } \quad yards\\ \\ \frac { 1 }{ 1760 } \quad miles\quad =\quad 1\quad yard$

You might be interested in

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago

An object is located 10.0 cm from a convex mirror. The magnitude of the

Sunny_sXe [5.5K]

Answer:

B. 6.00 cm

Explanation:

6 0

3 years ago

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

muminat

Answer:

The speed of the plank relative to the ice is:

$v_{p}=-0.33\: m/s$

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

$m_{g}v_{g}+m_{p}v_{p}=0$ (1)

Where:

m(g) is the mass of the girl
m(p) is the mass of the plank
v(g) is the speed of the girl
v(p) is the speed of the plank

Now, as we have relative velocities, we have:

$v_{g/b}=v_{g}-v_{p}=1.55 \: m/s$ (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

$45v_{g}+168v_{p}=0$

$v_{g}-v_{p}=1.55$

$v_{p}=-0.33\: m/s$

I hope it helps you!

8 0

3 years ago

A rocket is launched from the surface of Earth with a speed v0 that will allow the rocket to escape the gravitational field of E

weeeeeb [17]

Answer:

option ( a ) is correct .

Explanation:

Escape velocity on the earth = √ ( 2 GM / R )

where G is universal gravitational constant , M is mass of the earth and R is radius .

V₀ = √ ( 2 GM / R )

escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth

Radius of the planet = 4 R

escape velocity of planet = √ ( 2 GM / 4R )

= .5 x √ ( 2 GM / R )

= .5 V₀

option ( a ) is correct .

8 0

3 years ago

At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/

Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0

3 years ago