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Korvikt [17]
3 years ago
6

Electromagnetic force is present when electromagnetic fields

Physics
2 answers:
MAVERICK [17]3 years ago
6 0

Answer:

C- Interact with charged particles.

Explanation:

When a charged particle is placed in the field of the other charged particle then it will experience electromagnetic force on it.

aleksandr82 [10.1K]3 years ago
3 0

Answer: C; interact with charged particles.

Explanation: The elctromagnetic force is defined as:

F = q*E

Where q is the charge of a particle or charged object, and E is the electrical field, where the direction of the electric field defines the direction of the force.

Then, we need to have a charged particle in a outer eletrical field to have a electromagnetic force, which means that the correct answer is the option C; interact with charged particles.

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If a body of mass 5og moves in a circular path of radius 10cm .find work done
Sindrei [870]

Answer:

0Nm, no work is done.

Explanation:

Work done is defined as the Force per distance meaning force times the distance moved in the direction of the force.

Now the body of mass 50g has a centripetal force acting on it directed towards the centre. Now in actuality the body stays along the circle it doesn't really move to the centre of the circle.

Hence the force doesn't move a distance, and so from the definition of work done;

F×d ; d =0

Hence work done = mv2/r × 0= 0Nm

3 0
3 years ago
A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What
Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

a_{c} =\frac{v^{2} }{r}

Where a_{c} is the centripetal acceleration

v is the velocity

and r is the radius

From the given information

v = 12 \ m/s

and r = 30 \ m

Therefore,

a_{c} =\frac{12^{2} }{30}

a_{c} =\frac{144 }{30}

a_{c} = 4.8\ m/s^{2}

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

4 0
2 years ago
A glass rod is charged by the process of ________. *
Vedmedyk [2.9K]
Magnetism.

HOPE THIS HELPS !
7 0
3 years ago
Tell me any 5 importance of medicinal plants​
s344n2d4d5 [400]

Answer:

Aloe, Tulsi, Neem, Turmeric, and Ginger are medicinal plants that can help with a variety of diseases. Ginger, green tea, walnuts, aloe, pepper, and turmeric are just a few of these plants. Some plants and their derivatives are key sources of active compounds used in aspirin and toothpaste, among other things.

Explanation:

Plant name:                                    Uses:

1. Marshmallow:      ////  Relief from aching muscles and pain in muscle, Heals insect bite.             ////

2.California poppy ////  Relieves tension, Removes nervous system

3. Tulsi                ////   Cures sore throat, Cures fever and asthma

4. Neem               ////       Cures skin diseases, Cures diabetics

5. Aloevera      ////             Heals burns, Relieves constipation

(Hope this helps can I pls have brainlist (crown)☺️)

3 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
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