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4 years ago

12

A horse runs for 15 seconds to the south at a speed of 12 m/s. What is the

1 answer:

elixir [45]4 years ago

8 0

Answer: 180 m south

Explanation:

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You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abo

MakcuM [25]

Answer:

$v=87.46m/s$

Explanation:

Objects moving in circular path would be have either centripetal or centrifugal force.The force is either to center or away from center. When the object is moving along the circular path the centripetal force is

$F=\frac{mv^{2}}{r}$

Here m is mass, v is velocity and r is radius of circular path

The acceleration is given by:

$a_{r}=\frac{v^{2}}{r}$

The point of interest is lowest point on circle.The acceleration of plane at this position point up.The speed of plane from radial acceleration equation is:

$v=\sqrt{a.r}\\$

Substitute 17 m/s² for a and 450m for r

So

$v=\sqrt{17m/s^{2}*450m }\\ v=87.46m/s$

7 0

3 years ago

Which best describes accuracy?

Ierofanga [76]

The dependence of a measured value on a controlled value should be the correct answer.

7 0

3 years ago

Read 2 more answers

You are a nurse in a hospital. You are told to move a patient through three corridors each measuring 14.33 meters. Because of fr

dimulka [17.4K]

Work done by force is given by formula

$W = F.d$

now here work done against friction is given so force of friction here is

$F_f = 848 N$

It moved by three corridors with each measures

$d = 14.33 m$

so total distance will be

$s = 14.33 \times 3 = 43 m$

now from above formula of work done we will say

$W = 848 \times 43$

$W = 36464 J$

so above is the work done to move three corridors

3 0

3 years ago

Discuss the properties of a scalar quantity. give an example of a scalar.

lina2011 [118]

It has only magnitude of body and not indicate the direction of body.Example. electric potential force.

5 0

4 years ago

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear t

labwork [276]

Answer:

<h3>(A) The width $x = 0.24 \times 10^{-3}$ m</h3><h3>(B) The new width is $1.32 \times 10^{-3}$ m</h3>

Explanation:

Given :

Focal length $f = 135 \times 10^{-3} m$

Maximum aperture $D = \frac{f}{4}$

Wavelength $\lambda = 550 \times 10^{-9}$ m

(A)

From rayleigh criterion,

$\theta = \frac{1.22 \lambda }{D}$

$\theta =\frac{ 1.22 \times 550 \times 10^{-9} }{33.75 \times 10^{-3} }$

$\theta = 1.98 \times 10^{-5}$ rad

From angle formula,

$x = R\theta$

Where $R =$ 12 m ( given in example )

$x = 12 \times 1.98 \times 10^{-5}$ m

$x = 23.76 \times 10^{-5}$

$x = 0.24 \times 10^{-3}$ m

(B)

We know that $\theta$ is proportional to the $x$ and inversely proportional to the $D$

so we write the new width, here $x$ is 5.5 times larger than above case

$x = 0.24 \times 10^{-3} \times \frac{22}{4}$

$x = 1.32 \times 10^{-3}$ m

6 0

3 years ago