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3 years ago

What are the different kinds of forces?

Physics

2 answers:

Anon25 [30]3 years ago

6 0

Gravity, friction, and air resistance are some examples.

horsena [70]3 years ago

3 0

Friction, Tension, Gravitational, Electric, and magnetic

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A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi

zimovet [89]

Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

6 0

3 years ago

Along the remote Racetrack Playa in Death valley, California,stones sometimes gouge out prominent trails in the desert floor, as

MArishka [77]

Answer:

Explanation:

given ;

coefficient of kinetic friction = 0.80

mass m = 26kg

considering the force acting in horizontal direction and from newton's 2nd law of motion;

for vertical motion = Fn - mg = 0

for horizontal motion = F = ma + miu mg = m( a + miu.g)

but a = 0

therefore, F = miu mg where g = 9.81m/s^2

plugging the values into the equation;

F = 0.80 x 9.81 x 26

Horizontal force = 204.05N

3 0

3 years ago

To open a soda can lid, you can apply a force of 50 N to a car key wedged under

AysviL [449]

Answer:

7.8

Explanation:

Here, an effort of 50 N is applied at the car key and overcomes the resistance(or load) of 390 N at the lid.

mechanical advantage

=load/effort

=390 N/50 N

=7.8

8 0

3 years ago

Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven tra

Black_prince [1.1K]

Answer:

$\frac{F_1}{F_2}=3.55$

Explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

$v_2=\frac{2}{5}v_1$

$F_1=\frac{1}{2}\rho_1 CAv_1^2$

$F_2=\frac{1}{2}\rho_2 CAv_2^2\\\Rightarrow F_2=\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2$

Dividing the above two equations we get

$\frac{F_1}{F_2}=\frac{\frac{1}{2}\rho_1 CAv_1^2}{\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2}\\\Rightarrow \frac{F_1}{F_2}=\frac{\rho_1}{\rho_2\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=\frac{0.38}{0.67\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=3.55$

The ratio of the drag forces is $\mathbf{\frac{F_1}{F_2}}=\mathbf{3.55}$

5 0

3 years ago

In a posteroanterior (pa) projection of the chest being used for cardiac evaluation, the heart measures 14.7 cm between its wide

Ilya [14]

The actual diameter of the heart is 12.25 cm. Given : Heart measure = 14.7 cm Magnification factor = 1.2

14.7 / 1.2 = 12.25

8 0

3 years ago