Which tools would be necessary to determine whether or not a large block will float, without using water?

11) The circumstance MOST LIKELY to cause a compass to give an erroneous reading is standing too close to

Likurg_2 [28]

That would be the presence of a magnet anywhere near the compass.

3 0

3 years ago

Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom

Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

$B = B_o(-\hat k)$

now the velocity is from bottom to top

so we have

$v = v_o \hat j$

now the force on the moving charge is given as

$\vec F = q(\vec v \times \vec B)$

now we have

$\vec F = (-e)(v_o \hat j \times B_o(-\hat k))$

$\vec F = e v_o B \hat i$

so force will be towards Right

7 0

3 years ago

A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass

ValentinkaMS [17]

Answer:

Explanation:

a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis after explosion of third part = mv cosθ

= 10 v cosθ

Momentum along x of first part = - 5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

v cosθ = 21

Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 + 10 v sinθ

v sinθ= 21

squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

v² = 441 + 361

v = 28.31 m/s

Tanθ = 21 / 19

θ = 48°

6 0

3 years ago

Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface

salantis [7]

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

R = 2r

= $2 \times 0.1 m$

= 0.2 m

where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

U = $\frac{k \times Q \times Q}{R}$

= $\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}$

= 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, $\frac{U}{2}$ .

Hence, K.E = $\frac{U}{2}$

= $\frac{0.081}{2}$

= 0.0405 J

Now, we will calculate the speed of balls as follows.

V = $\sqrt{\sqrt{\frac{2 \times K.E}{m}}$

= $\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}$

= 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

6 0

2 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

Andrei [34K]

Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

7 0

3 years ago