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2 years ago

John pushes his brother Danny on a skateboard. John applies a force of 29 N and Danny's acceleration is 0.4 m/s^2

Physics

2 answers:

erastova [34]2 years ago

7 0

The combined mass of Danny+skateboard is 72.5 kg

Explanation:

We can solve the problem by using Newton's second law of motion:

$F=ma$

where:

F is the net force applied on Danny and the skateboard

m is the mass of Danny+skateboard

a is the acceleration

Here we have the following data:

F = 29 N is the force applied

$a=0.4 m/s^2$ is the acceleration of Danny and the skateboard

Solving for m, we find their combined mass:

$m=\frac{F}{a}=\frac{29}{0.4}=72.5 kg$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

Alexxandr [17]2 years ago

7 0

Answer:

73 kg A P E X answer

Explanation:

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Answer: Increase in wave frequency

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2 years ago

A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met

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Answer:

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Explanation:

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In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

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Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

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2 years ago

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

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Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

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2 years ago