1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
san4es73 [151]
2 years ago
9

Five properties of magnet​

Physics
1 answer:
Olenka [21]2 years ago
3 0

Answer:

1. The magnet is magnetic and can attract iron articles.

2. The magnet has magnetic poles. Each magnet has two kinds of poles: N pole and S pole. They are in pairs.

3. Temporary magnet and permanent magnet: when the ferromagnetic material is magnetized, it is easy to lose the magnetic property, which is called temporary magnet (for example: iron); when the ferromagnetic material is magnetized, it is not easy to lose the magnetic property, which is called permanent magnet (for example: steel).

4. When two magnets are close to each other, the same poles will repel and push away from each other, and the different poles will attract and stick to each other. Therefore: the same pole repels each other, the different pole attracts each other.

5. The attraction of a magnetic object is called magnetism. An object is surrounded by a magnetic material. The area affected by the magnetic force is called the magnetic field.

You might be interested in
A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0
3 years ago
During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the
natita [175]

Answer:

Explanation:

If the work done on the cart is NET work

Then the work will result in an increase in kinetic energy

KE₀ + W = KE₁

½mv₀² + W = ½mv₁²

½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²

v₁ = 1.626991...

v₁ = 1.6 m/s

4 0
2 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

7 0
3 years ago
Question 5 (2 points)
Radda [10]

Answer:

Increase the work being done or decrease the time in which the work is completed

Explanation:

I got it right on the quiz i just took :)

5 0
3 years ago
Other questions:
  • Give a combination of four quantum numbers that could be assigned to an electron occupying a 5p orbital.
    13·1 answer
  • If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the
    12·1 answer
  • Is diet Pepsi heterogeneous or is it homogeneous
    8·2 answers
  • During times of stress, Yanira notices that her heart pounds a little faster than normal. What part of the brain is responsible
    6·2 answers
  • Light and radio waves travel through a vacuum in a straight line at a speed of very nearly 3.00 ´ 108 m/s. How far is light year
    14·1 answer
  • a pennyfarthing is a style of a bicycle with a very large front wheel and a small real wheel, the cyclist who sit high above and
    10·1 answer
  • What does a model of a light wave tell us about<br> brightness and color?
    6·1 answer
  • a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev
    11·1 answer
  • All of the following are dimensions of progressive overload except multiple choice specificity. time. intensity. frequency.
    12·1 answer
  • Even though it's not very popular these days, I like our system . Getting together to talk about who the best candidate might be
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!