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3 years ago

If you blow harder into a trumpet you will create a sound wave with a greater

Physics

2 answers:

Furkat [3]3 years ago

5 0

Answer:

C. wavelength

<em>good luck, i hope this helps :)</em>

ElenaW [278]3 years ago

4 0

Answer is C

Explanation:

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A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$

Now, differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}$

Now, we are given speed as 25 mph. So, $\frac{de}{dt}=25\ mph$

Also, when $e=91\ mi$ , we can find 'x' using equation (1). This gives,

$x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi$

Now, plug in the values of 'e' and 'x' and solve for $\frac{dx}{dt}$ . This gives,

$\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph$

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0

3 years ago

A bus travels 30km in 1/2h. What is it's average speed in km/h

Vadim26 [7]

If a bus travels 30 km in 1/2 hr, then in one hr, he can travel twice the distance.
30*2=60 km

Final answer: 60 km per hr

5 0

3 years ago

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

3 years ago

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = $\frac{kQ_{1}Q_{2} }{r^{2} }$

F= $\frac{K(Q/2)(3Q/4)}{r^{2} }$

how $\frac{KQ^{2} }{r^{2} }$ = 0.42

Then

F = $\frac{0.42*3}{8}$

F = 0.1575 N

3 0

3 years ago

A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. (See Figure 1) However, he is unable t

cupoosta [38]

Answer:

Normal force, $N = W_s - F$

Explanation:

Let $w_s$ is the weight of suitcase. A man attempts to pick up his suitcase by pulling straight up on the handle. The weight of the suitcase in downward direction. The normal force is acting in upward direction. Let F is the force with which it is pulled straight up.

So, the normal force is given by :

$N = W_s - F$

N = mg - F

mg is the weight of the suitcase.

Hence, this is the required solution.

3 0

3 years ago