A line dance that is fast in pace will most likely:

Witch of the following is the best exsample of potential energy

An object or living organism that is stationary(not moving) for example the car is stopped at the top of the hill

8 0

3 years ago

With light microscopy, if the objective lens (lens closest to the specimen) magnifies 40-fold, and the eyepiece lens magnifies 1

Fed [463]

The final magnification will be 400-fold or 400 times the original size of the object.

For magnifying smaller objects, a compound microscope is used.

A compound microscope consists of an objective and an eyepiece, whose diagram is shown in the adjoining image.

The lens near the object is called an objective and the other one is the eyepiece.

Let the magnification of the objective be m1

Let the magnification of the eyepiece be m2

The final magnification by the microscope, M, will be

M = m1 x m2

Putting the values in the above equation

M = 40 x 10

M= 400

Thus, the final magnification will be 400-fold or 400 times the original size of the object.

To know more about "optical instruments", refer to the link given below:

brainly.com/question/13276240?referrer=searchResults\

#SPJ4

6 0

2 years ago

If mass remains the same while the volume of a substance ________, the density of the substance will_______________. decreases,

gtnhenbr [62]

Density =Mass/volume
When mass is constant, density is inversely proportional to volume or vice versa
When volume decreases, density increases
Hope it helps

7 0

3 years ago

Read 2 more answers

An asteroid has a radius of 110,000 m

nydimaria [60]

Answer: M = 6.13 × 10^18 kg

Explanation:

g = GM/r2,

Where

The mass M of the asteroid = ?

The radius r = 110000 m

g = 0.0338 m/s^2

G is the gravitational constant.

SI units its value is approximately 6.674×10^−11m3⋅kg−1⋅s−2

Using the formula

g = GM/r2

Cross multiply

GM = gr^2

6.674×10^-11M = 0.0338 × 110000^2

M = 408×10^6/6.674×10^-11

M = 6.13 × 10^18 kg

7 0

4 years ago

A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total

NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

(c) Magnitude of the average speed is given by:

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$

6 0

3 years ago