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Marta_Voda [28]
3 years ago
6

An alpha particle travels at a velocity of magnitude 550 m/s through a uniform magnetic field of magnitude 0.045T. (An alpha par

ticle has a charge of charge of +3.2 x 10-19 C and a mass 6.6 x 10-27 kg) The angle between the particle's direction of motion and the magnetic field is 52°. What happens to the speed? The answer is that the speed remains the same and I want to know why and how would we know if the speed changes ( there are two questions before this about the force and acceleration force is 6.2x10^-18 and acceleration is 9.5x10^8 )
Physics
1 answer:
creativ13 [48]3 years ago
3 0

Answer:

the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Explanation:

The magnetic force is given by the expression

    F = q v x B

The bold are vectors, where v is the velocity and B is the magnetic field, the product is the cross product whose result is a vector perpendicular to the two vectors (v and B)

From the above, the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

   

Even when the change in direction is real and is caused by a centripetal force

For there to be a change in the velocity modulus there must be a force parallel to the velocity direction, generally a force in electrical

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13. An object with a mass of 2.0 kg has a force of 4.0 newtons applied to it.
julsineya [31]

Answer:2m/s²

Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values

4N=2KG*A

A=4N/2KG

A=2m/s²

7 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
1 year ago
Please help and check all that apply and I will mark brainliest if it’s correct
Yuri [45]
A syncline is visable
3 0
3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
2 years ago
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