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3 years ago

a car travel the first 20km with a speed of 40km/h and the next 40km with a speed of 80km/h . find the average speed

Physics

1 answer:

REY [17]3 years ago

3 0

Answer:

average speed is 60km/h

Explanation:

you sum up the speed attained in each distance covered and divide it by 2 to get your answer

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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

1. The following can be inferred from Newton’s second law of motion except:

lora16 [44]

Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)

2.a)acceleration decreases

Explanation:

Newton's second law:

Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.

When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.

This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .

7 0

3 years ago

How are light waves used to bring far away objects into view and how does the eye translate them?

adelina 88 [10]

Answer:

When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.

Explanation:

Hope it will help u

5 0

3 years ago

Explain the role of good marketing in the success of business

Akimi4 [234]

Answer:

I am joker if you are mad x_________-..............,,,,,,........ €£

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3 years ago

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0

3 years ago

a car travel the first 20km with a speed of 40km/h and the next 40km with a speed of 80km/h . find the average speed​

a car travel the first 20km with a speed of 40km/h and the next 40km with a speed of 80km/h . find the average speed