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2 years ago

What can be used to plot the magnetic field around a bar magnet?

Physics

1 answer:

zlopas [31]2 years ago

4 0

Answer:

A bar magnet creates a magnetic field around it. The field cannot be seen, but its effects can be observed. This is how the magnetic field around a magnet is detected. Iron fillings can be used to study the presence of magnetic field and a compass can be used to plot the field lines.

Explanation:

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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago

Where does most of earths available carbon come from

cupoosta [38]

Rocks and sediments I believe

6 0

3 years ago

A wide sloping deposit of sediment formed where a stream leaves a mountain range is called

tatuchka [14]

An alluvial fan is a wide, sloping deposit of sediment formed where a stream leaves a mountain range. Make sure not to confuse it with a delta. A delta is a landform made of sediment that is deposited where a river flows into an ocean or lake. Hope this helped!

7 0

3 years ago

The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel

IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

Complete question

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

$W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2} )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\$

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0

3 years ago

A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86

Vladimir [108]

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

3 0

3 years ago