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Nataly_w [17]
2 years ago
13

What do the magnetic field lines on a bar magnet look like?

Physics
2 answers:
KATRIN_1 [288]2 years ago
6 0

Answer

A is the correct answer

Explanation:

mark me as brainlist and follow me

zubka84 [21]2 years ago
3 0

Answer:

D

Explanation:

On a bar magnet, the lines start at the north pole and loop around to the south pole.  A diagram is attached for any further clarifications.

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Which of the following is the correctly abbreviated SI unit describing the mass of an object?
Aloiza [94]

The correct answer would be kg!

6 0
3 years ago
If steam enters a turbine at 600°K and is exhausted at 200°K, calculate the efficiency of the engine.
alekssr [168]
Answer: e = 3.333% or 33 1/3%
                       ^
                  repeated

Identify the given information.
Work input (Wi)= 600 K
Work output (Wo) = 200 K

Choose which formula you should use. In this case, you are finding efficiency.
e =  (Wo/Wi) x 100%

Substitute and solve. 
e = (200 K/600 K) X 100%
e = 0.333 X 100%
e = 33.333% or 33 1/3%
5 0
3 years ago
Read 2 more answers
A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for
Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0
4 years ago
A raging bull of mass 700kg runs at 10 m/s. How much kinetic energy does it have?
Illusion [34]

Answer: 35000 J or 35kJ

Explanation:

Use equation for kinetic energy : Ek=mV²/2

m=700 kg

V=10m/s

-------------------

Ek= 700kg*100m²7s²/2

Ek=35000 J=35kJ

6 0
3 years ago
A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
Vika [28.1K]

Answer:

t_t=4.131\ s

Explanation:

Given:

height above the horizontal form where the ball is hit, y=1\ m

angle of projectile above the horizontal, \theta=30^{\circ}

initial speed of the projectile, u=40\ m.s^{-1}

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

  • v_y=0\ m.s^{-1}

Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

h=20.4082\ m (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

  • we've total height, h'=h+y =20.4082+1

h'=u_y'.t'+\frac{1}{2} g.t'^2

  • during the course of descend its initial vertical velocity is zero because it is at the top height, so u_y'=0\ m.s^{-1}

21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

t_t=4.131\ s

3 0
3 years ago
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