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horsena [70]
3 years ago
13

Compasses line up with magnetic fields. A compass will line upA. Parallel to magnetic field lines, with the south pole pointing

in the direction of the field.B. Perpendicular to magnetic field lines.C. Parallel to magnetic field lines, with the north pole pointing in the direction of the field.D. Perpendicular to magnetic field lines, in a direction defined by the right-hand rule.
Physics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

C. Parallel to magnetic field lines, with the north pole pointing in the direction of the field.

Explanation:

When a magnet is placed in magnetic field then its North pole and south pole will experience magnetic force in opposite directions.

North pole will tend to move in the direction of magnetic field while south pole has tendency to move opposite to it.

Due to this opposite direction force magnet will experience torque if it is inclined at some angle with magnetic field

Due to this torque magnet will align in the direction of magnetic field such that its north pole will point in the direction of external field while south pole is align opposite to that

Olenka [21]3 years ago
4 0

you are welcome i hope god blesses you today

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If you use a force of 90 N to pick up a 10 pound bag of charcoal, what is the acceleration?
hjlf

Answer:

9ms^2

Explanation:

since ,Force=mass*acceleration

then, acceleration=force/mass

and, Force=90N

Mass=10pound

therefore, acceleration=90/10

=9ms^2

8 0
3 years ago
A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H
Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

3 0
3 years ago
A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law and notes that the spring stretches a d
Irina-Kira [14]

Answer:

a) 0.040625 m

b) 5.02272 J

Explanation:

k = Spring constant

x = Stretched length

F = Force

a)

F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{4\times 9.81}{0.025}\\\Rightarrow k=1569.6\ N/m

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{6.5\times 9.81}{1569.6}\\\Rightarrow x=0.040625\ m

Extension of the spring would be 0.040625 m

b) Work done in a spring

W=\frac{1}{2}kx^2\\\Rightarrow W=\frac{1}{2}\times 1569.6\times 0.08^2\\\Rightarrow W=5.02272\ J

The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J

4 0
3 years ago
A 15.0 Kg object is moved from a height of 7.00 m above thefloor to a height of 13.0 m above the floor. What is the change ingra
Pie

To solve this problem we will apply the concepts related to potential gravitational energy. This is defined as the product between mass, acceleration and change in height and can be expressed as,

\Delta PE = mg \Delta h

Here,

m = Mass

g = Gravitational acceleration

\Delta h = Height

Replacing with our values we have,

\Delta PE = (15kg)(9.81m/s^2)(13m-7m)

\Delta PE = 882.9J \approx 883J

Therefore the change in gravitational potential energy is 883J.

4 0
3 years ago
A 75-turn coil with a diameter of 6.00 cm is placed in a constant, uniform magnetic field of 1.00 T directed perpendicular to th
kolezko [41]

Answer:

The induced emf in the coil at the t = 5s is 6.363 mV

Explanation:

Given;

number of turns = 75

diameter of the coil = 6 cm

magnetic field strength = 1 T

new magnetic field strength = 1.30 T at t = 10.0 s

Area \ of \ coil = \frac{\pi d^2}{4} =  \frac{\pi *0.06^2}{4} = 0.002828 \ m^2

E.M.F = \frac{NA* \delta B}{\delta t}

Between 0 to 5 s, Induced emf is given as;

E.M.F = \frac{75*0.002828*(B_5-1)}{5}

Between 5 to 10 s, Induced emf is given as;

E.M.F = \frac{75*0.002828*(1.3-B_5)}{5}

Since the field increased at a uniform rate until it reaches 1.30 T at t = 10.0 s, the induced emf will also increase in uniform rate. And equal time interval will generate same increase in field strength.

B₅ -1 = 1.3 - B₅

2B₅ = 2.3

B₅ = 1.15 T

Thus, magnetic field at t = 5 is 1.15 T

E.M.F = \frac{75*0.002828*(1.3-1.15)}{5} = 6.363 \ mV

Therefore, the induced emf in the coil at the t = 5s is 6.363 mV

7 0
3 years ago
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