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BARSIC [14]
3 years ago
6

. A child pulls a sled up a snow-covered hill. In the process, the child does 405 J of work on the sled. If she walks a distance

of 15 m up the hill, how large a force does she exert on the sled?
Physics
1 answer:
leonid [27]3 years ago
5 0
<span>Work done by child W = 405 J; Distance walked s = 15 m
 Calculating the force exerted on the sled,
 Work = Force x distance
 => Force F = Work W / distance s => => F = 405 / 15 = 27 N
 Force F is 27 Newtons.</span>
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The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
4 years ago
Please help will give the brainliest<br><br> How did humans decide how long a year should be?
worty [1.4K]

Answer:

The decided by staring at and observing the stars.

Explanation:

5 0
3 years ago
Read 2 more answers
Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0
Sati [7]

Answer:

Ex=  -17.1 N/C

Ey =  +26.9 N/C

Explanation:

We apply formula of electric field:

Ep=k*q/d²

Ep:  Electric field at point ( N/C)

q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1

Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)

Calculation of the field at point P due to the load q₂

d_{2} = \sqrt{1.30^{2}+0.450^{2}  }

d₂=1.375 m

E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂

Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β

, then,E₂ tiene componentes x-y  en P.

E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C

E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C

Calculation of the electric field at point P located at the origin(0,0)

Ex=E₂x= -17.1 N/C

Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C

4 0
3 years ago
the declaration of Independence discusses the protection of natural rights. These rights include the right life liberty and the
nikdorinn [45]

Answer:

i can't understand ur question

4 0
3 years ago
Question 2 (4 points) The cause of heat cramps is the loss of sodium and potassium through heavy sweating? True False ​
sattari [20]

Answer:

True

Explanation:

Lose of water content in our body through sweating and lose of essential salts such as potassium and sodium causes heat cramps...

Thank you

8 0
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