2. Connect concepts on the left with a description on the right.

What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?

KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

6 0

2 years ago

How do you calculate elastic potential energy

Nimfa-mama [501]

U=1/2kx2

This image sums it up

5 0

3 years ago

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>

The time period of an oscillation refers to the time taken by an object to complete one oscillation.
It is the inverse of frequency of oscillation; denoted by "T".

Now,

$\sqrt{\frac{L}{g}}$ , where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
The period will increase as one climbs a very tall mountain because g will slightly decrease.
Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

#SPJ4

6 0

1 year ago

The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov

son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The eyepiece focal length is $f_e = 0.027 \ m$

Explanation:

From the question we are told that

The focal length is $f_o = 5.5 \ mm = -0.0055 \ m$

This negative sign shows the the microscope is diverging light

The angular magnification is $m = 300$

The distance between the objective and the eyepieces lenses is $Z = 19 \ cm = 0.19 \ m$

Generally the magnification is mathematically represented as

$m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]$

Where $f_e$ is the eyepiece focal length of the microscope

Now making $f_e$ the subject of the formula

$f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }$

substituting values

$f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }$

$f_e = 0.027 \ m$

5 0

2 years ago

In a classroom demonstration, the pressure inside a soft drink can is suddenly reduced to essentially zero. Assuming the can to

umka2103 [35]

Answer:

3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.

Explanation:

The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =

A = 2πr(r + h)

Where h = height of the can = 12cm

r = radius of the can = 6.5cm/2 = 3.25cm

r = diameter /2

A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²

Atmospheric pressure, P = 101325Pa = 101325 N/m²

F = P × A

F = 101325 ×0.031.

F = 3141N. Or 3.1 ×10³ N.

5 0

3 years ago