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3 years ago

A wooden cylinder has a density of 4 g/cm3 and a mass of 48 cm3. What is the volume of the wooden cylinder?

Physics

1 answer:

Tema [17]3 years ago

4 0

Answer:

Volume is 12 cm³

Explanation:

$volume = \frac{mass}{density} \\$

mass » 48 g

density » 4 g/cm³

$volume = \frac{48}{4} \\ \\ = 12 \: {cm}^{3}$

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creativ13 [48]

Answer:

i dont understand, can you put it in word form please.

Explanation:

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3 years ago

Will the body be travelling with a constant speed or a constant acceleration if the distance travelled by the body is directly p

Anna35 [415]

Distance ( say d ) is directly proportional to time ( say t ) .
mathematically ,
d = kt
where k is constant if proportionality .
Now differentiating the above equation with respect to time ,
we get ;
v = k
=> velocity is constant .
=> No acceleration .

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4 years ago

An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If th

stepan [7]

Answer:

Number of photons travel through pin hole= $6.4*10^{17}$

Explanation:

First we will calculate the energy of single photon using below formula:

$E=\frac{h*c}{λ}$

Where :

h is plank's constant with value $6.626*10^{-34} J.s$

c is the speed of light whch is $3*10^{8}$

λ is the wave length = 532nm

$E=\frac{6.6268*10^{-34}* 3*10^{8}}{532nm}$

E= $3.73*10^{-19}$ J

Number of photons emitted per second:

$\frac{5J/s}{1 photon/3.73*10^{-19} }$

Number of photons emitted per second= $1.34*10^{19} photons/s$

$\frac{A-hole}{A-beam}$ = $\frac{\frac{pie*d-hole^2}{4}}{\frac{pie*d-beam^2}{4} }$

Where:

A-hole is area of hole

A-beam is area of beam

d-hole is diameter of hole

d-beam is diameter if beam

$\frac{A-hole}{A-beam}$ = $\frac{d-hole^2}{d-beam^2}$

$\frac{Ahole}{A-beam}$ = $\frac{1.22^2}{5.5^2}$

$\frac{A-hole}{A-beam}$ = $\frac{144}{3025}$

Number of photons travel through pin hole= $1.34*10^{19} *\frac{144}{3025}$

Number of photons travel through pin hole= $6.4*10^{17}$

7 0

3 years ago

A bicycle odometer which counts revolutions and is calibrated to report distance traveled is attached near the wheel axle and is

Bess [88]

Answer:

each rotation of the smaller wheel will show 84.382-75.3982=8.9838 inches more than the actual distance

Explanation:

d = Diameter of the wheel

The distance traveled in one rotation of the wheel is the circumference of the wheel

$\pi d=\pi\times 27\\ =84.823\ inch$

When diameter is 24 inches

$\pi d=\pi\times 24\\ =75.3982\ inch$

Therefore, each rotation of the smaller wheel will show 84.382-75.3982=8.9838 inches more than the actual distance

6 0

3 years ago

A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20

statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

Assuming no friction, total mechanical energy must keep constant, so the following is always true:

$\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)$

Choosing the ground level as our zero reference level, Uf =0.
Since the child starts from rest, K₀ = 0.
From (1), ΔU becomes:
$\Delta U = 0- m*g*h = -m*g*h (2)$
In the same way, ΔK becomes:
$\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)$
Replacing (2) and (3) in (1), and simplifying, we get:

$\frac{1}{2}*v_{1}^{2} = g*h (4)$

In order to find v₁, we need first to find h, the height of the slide.
From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

$h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)$

Replacing (5) in (4) and solving for v₁, we get:

$v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)$

As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
We can do this in more than one way, but a very simple one is using kinematic equations.
If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

$v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)$

Since v₀ = 0 (the child starts from rest) we can solve for a:

$a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)$

Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

$t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)$

Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

$x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)$

Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

$x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)$

From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

$v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)$

Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

$\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)$

Replacing from (12) in (13), we can solve for h₂:

$h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)$

Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

$x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)$

4 0

3 years ago