All categories

3 years ago

A bus travels 6 km east and then 8 km south. The magnitude of the bus’s resultant displacement is ___km.

Physics

1 answer:

Crank3 years ago

3 0

Answer:

10 km

Explanation:

You can draw your problem out. You can use the pathagoran theorem to figure it out. So $6^{2}$ + $8^{2}$ and then square root the answer which will be $\sqrt{100}$ which equals to 10.

You might be interested in

What is the purpose of the thermal energy tranfer lab

iren2701 [21]

Answer:thermal Energy Transfer in Mixtures Purpose: The purpose of this experiment is to discover how exactly the final temperature of a mixture, involving a substance and hot water, is affected and impacted by the type of substance used. This means that when hot water is mixed with another substance, it must be determined

Explanation:

did this on edge

6 0

2 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0

3 years ago

An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

jenyasd209 [6]

Answer:

a) i = -9.63 cm , h ’= .0.24075 cm erect

b) i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

1 / f = (n-1) (1 / R₁ - 1 / R₂)

1 / f = (1.70 -1)) 1 / ∞ - 1/13)

1 / f = 0.0538

f = - 18.57 cm

Now we can use the constructor equation

1 / f = 1 / o + 1 / i

1 / i = 1 / f - 1 / o

1 / i = -1 / 18.57 -1/20

1 / i = -0.1038 cm

I = -9.63 cm

For the height of the

image let's use magnification

m = h '/ h = - i / o

h ’= -h i / o

h ’= - 0.5 (-9.63) / 20

h ’= .0.24075 cm

b) we invert the lens

The focal length is

1 / f = (1.70 -1) (1/13 - 1 / int)

1 / f = 0.0538

f = 18.57 cm

1 / i = 1 / f -1 / o

1 / I = 1 / 18.57 - 1/20

1 / I = 3.85 10-3

i = 259.74 cm

h ’= - 0.5 259.74 / 20

h ’= 6.4935 cm

7 0

3 years ago

How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s

GenaCL600 [577]

We need more evidence to be provided

4 0

2 years ago

Which example best describes the term carrying capacity?

mezya [45]

When people aboard a plane...the amount of baggage you take has to vary because the plane has a certain carrying capacity.

3 0

3 years ago