According to the passage, why don’t sharks sink in the water?

Step2247 [10]

It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).

5 0

3 years ago

In a mass spectrometer, a single-charged particle (charge e) has a speed of 1.0 × 10 6 m/s and enters a uniform magnetic field o

Nonamiya [84]

Answer:

The mass is $m =6.4*10^{-28} \ kg$

Explanation:

From the question we are told that

The speed of the charge is $v = 1.0 *10^{6} \ m/s$

The magnetic field is $B = 0.20 \ T$

The radius is $r = 0.02 \ m$

The value of the charge is $e = 1.60 *10^{-19} \ C$

The centripetal acting on the charge moving in the circular orbit is mathematically represented as

$F_c = \frac{mv^2}{r }$

Now this centripetal force is due to the force exerted on the charge by the magnetic field on the charge which is mathematically represented as

$F_m = qv B sin\theta$

At the maximum of this magnetic force $\theta = 90 ^o$

So

$F_m = e v B sin(90)$

$F_m = e v B$

Now given that it is this magnetic force that is causing the circular motion we have that

$F_c = F_m$

=> $\frac{mv^2}{r } = ev B$

=> $m = \frac{e * B * r }{v }$

substituting values

$m = \frac{ 1.60 *10^{-19} * 0.20 * 0.020 }{1.0*10^{6} }$

$m =6.4*10^{-28} \ kg$

8 0

4 years ago

A 5.0kg toolbox is raised from the ground by a rope. If the upward acceleration of the bucket is 2.5 m/s^2, find the force exert

swat32

Answer:

62 N

Explanation:

Sum of the forces on the toolbox:

∑F = ma

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5.0 kg) (9.8 m/s² + 2.5 m/s²)

T = 61.5 N

Rounded to two significant figures, the force exerted by the rope is 62 N.

8 0

4 years ago

A tightly wound 1000-turn toroid has an inner radius 1.00 cm and an outer radius 2.00 cm, and carries a current of 1.50 A. The t

V125BC [204]

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

<u>Explanation:</u>

Given;

Number of turns, N = 1000

Inner radius, r₁ = 1cm

Outer radius, r₂ = 2cm

Current, I = 1.5A

Magnetic field strength, B = ?

The magnetic field inside a tightly wound toroid is given by B = μ₀ NI / 2πr

where,

a < r < b and a and b are the inner and outer radii of the toroid.

The magnetic field of toroid is

$B = \frac{u_oNI}{2\pi r}$

Substituting the values in the formula:

$B (1.10cm) = \frac{(4\pi X 10^-^7 ) ( 1000)(1.5)}{2\pi (1.10) } \\\\$

$B (1.10cm) = 27.3mT$

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

7 0

3 years ago

a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat

lukranit [14]

1) The minimum force needed is 920 N

2) The minimum force is 460 N

3) The minimum force is 598 N

Explanation:

1)

We can answer this part by simply looking at the forces involved. In fact, there are two forces acting on the railroad:

Its weight, W, acting downward
The force applied to lift it, F, upward

So the net force on the railroad is

$\sum F = F - W$

where

W = 920 N is the weight of the railroad

In order to lift the railroad, the net force must be upward, so

$\sum F \geq 0$

And therefore

$F\geq W$

which means that the minimum force needed is equal to the weight of the railroad, 920 N.

2)

In this case, we have to use the principle of equilibrium of moments.

In fact, when the railroad rotates uniformly (=constant angular speed) about its end, it means that the moment produced by the weight (acting in one direction) is equal to the moment produced by the force applied (acting in the other direction). Therefore, we can write:

$W \frac{L}{2} = F L$

where

W = 920 N is the weight

L = 2.6 m is the length of the railroad

F is the force applied

We wrote L/2 on the left of the equation because the weight acts at the center of mass of the railroad (located at the midpoint), while on the right it is L because the force F is applied at the end of the railroad.

Solving for F,

$F=\frac{W}{2}=\frac{920}{2}=460 N$

3)

This problem is similar to the previous part, however in this case, the force applied F is applied 0.6 m from the end, pivoting around the opposite end.

This means that the distance between the point of application of the force F and the pivot is

L' = L - 0.6

where

L = 2.6 m

Therefore the equation for the equilibrium of moments becomes

$W\frac{L}{2}=F(L-0.6)$