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3 years ago

10

In an orbiting space shuttle, you are handed two identical boxes, one filled with sand and the other filled with feathers. How c

an you determine which is which without opening the boxes?

1 answer:

Aleks04 [339]3 years ago

3 0

Explanation:

Simply, shake the boxes. The box that is difficult to shake will be the sand and the box that will be easier will be the feathers.

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Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati

jek_recluse [69]

Answer:

$a=4\ m/s^2$

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2$

So, the magnitude of centripetal acceleration is $4\ m/s^2$ .

5 0

2 years ago

Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi

Leviafan [203]

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

<u>W = 12.96 J</u>

5 0

3 years ago

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide

Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is $4.2\ cm$

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

$v=\sqrt{\dfrac{2eV}{m}}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}$

$v=15689290.81\ m/s$

$v=1.56\times10^{7}\ m/s$

(b). We need to calculate the radius of electron

Using formula of centripetal force

$\dfrac{mv^2}{r}=qvB$

$r=\dfrac{mv}{qB}$

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}$

$r=0.042\ m$

$r=4.2\ cm$

Hence, (a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is 4.2 cm

8 0

3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

3 years ago

State two factors that affect the type of Shadows formed

Vadim26 [7]

Answer:

The two factors that affect the type of Shadows formed are stated below ::

The distance between the source of land and the object.
The angel where the ray of light falls on the object.

7 0

2 years ago

Read 2 more answers