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Sergio [31]
3 years ago
13

Determine the magnitude of the resultant force FR=F1+F2FR=F1+F2. Assume that F1F1F_1 = 235 lblb and F2F2F_2 = 350 lb

Physics
1 answer:
Svetlanka [38]3 years ago
4 0

Answer:

<em>585lb</em>

Explanation:

Given the formula for calculating the magnitude of the resultant force as;

F_R = F_1 + F_2

<em>Given </em>

<em></em>F_1 = 235 lb \ and \ F_2 = 350 lb\\F_R = 235lb + 350lb\\F_R = 585lb\\<em></em>

<em>Hence the magnitude of the resultant force is 585lb</em>

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stepan [7]

Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.

<h3>What is radiative diffusion?</h3>

A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.

As photons, energy passes through the radiation zone as electromagnetic radiation.

The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.

Hence,radiative diffusion is correct answer.

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7 0
2 years ago
PLEASE help with this question!
Ainat [17]

Answer:

the answer might the number 2

Explanation:

6 0
2 years ago
Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of
Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}

m is mass of Singly charged uranium-238 ion, m=3.95\times 10^{-25}\ kg

q is charge

So,

r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m

So, the radius of their circular path is equal to 0.0548 m.

4 0
3 years ago
A body weighs 100newtons and 80newtons when submerged in water.calculate the upthrust action on the body
katovenus [111]

Answer:

20 N

Explanation:

In air, the normal force is equal to the weight.

∑F = ma

N − mg = 0

N = mg

Submerged in water, the normal force is equal to the weight minus the buoyant force:

∑F = ma

B + N − mg = 0

N = mg − B

Plugging in values:

80 N = 100 N − B

B = 20 N

3 0
3 years ago
5.3 two 30 kg children in a 20 kg cart are stationary at the top of a hill. They start rolling down the 80 m tall hill and they
Makovka662 [10]

Answer:

<em>60008.4 J</em>

<em></em>

Explanation:

The mass of each kid = 30 kg

mass of the cart = 20 kg

The speed of the cart down the hill = 30 km/hr = 30 x 1000/3600 = 8.33 m/s

The height of the hill = 80 m

The potential energy of the boys at the top of the hill = mgh

where

m is the total mass of the kids and the cart = (30 x 2) + 20 = 80 kg

g is the acceleration due to gravity = 9.81 m/s^2

h is their height above the ground = 80 m (on the top of the hill)

substituting, we have

potential energy PE = 80 x 9.81 x 80 = 62784 J

At an instance at the bottom of the hill

their kinetic energy = \frac{1}{2} mv^2

where

v is their velocity = 8.33 m/s

m is their total mass = 80 kg

substituting, we have

kinetic energy KE = \frac{1}{2}*80*8.33^2 = 2775.6 J

Total work done on the cart is equal to the energy lost by the cart when it reached the bottom of the hill

work done by friction = PE - KE = 62784 - 2775.6 = <em>60008.4 J</em>

5 0
3 years ago
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