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d1i1m1o1n [39]
3 years ago
5

Akbar and Lucia apply opposing forces on a

Physics
1 answer:
iragen [17]3 years ago
3 0

Answer:

20 N

Explanation:

We can solve this problem by applying Newton's second law of motion, which states that the net force applied on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass

a is the acceleration

In this problem, we have:

m = 10 kg is the mass of the cart

a=2 m/s^2 is the acceleration

So the net force on the car is

F=(10)(2)=20 N

Since Lucia and Akbar are pushing into opposite directions, this means that this net force is the resultant of the two forces: and since the cart is accelerating towards Akbar, this means that Lucia is applying a force 20 N stronger than Akbar.

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a wave in the ocean has an amplitude of 3m. Winds pick up suddenly, increasing the wave's amplitude to 6m. How does this change
Paha777 [63]

Answer

Hi,

An increase in amplitude from 3m to 6 m increases the energy it transports. The frequency of the wave is not affected

Explanation

Amplitude is the height of a wave where as frequency is the number of waves that pass by each second. A wave with bigger amplitude has more energy than a wave with smaller amplitude. A point where more waves pass contains more energy that is transferred every second. The change in the amplitude of a wave does not change its frequency. However, frequency is inversely related to the wavelength of a wave.

Best Wishes!

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2 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

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