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3 years ago

How do you calculate the volume and density of something that weighs 20 grams?

1 answer:

6 0

Mass divided by volume is density, while mass times density is volume. you cannot calculate density without volume and you cannot calculate volume without density.
i believe that's the answer.
oh gosh i didn't realize my middle school education involved high school stuff.

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A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc

Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

1/f = 1/di + /1do

then the image distance

di = fd_o / d_o - f

= (10)(40) / 40-10

= 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

di = 10×20/ 20-10

= 20

Hence, the image must be real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 = ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0

2 years ago

Read 2 more answers

A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal

suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

$m_c$ = The center of mass of the stick = $\frac{L}{2}-0.22=0.5-0.22=0.28\ m$

$\omega$ = Angular velocity

Moment of inertia of the system is given by

$I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)$

As the energy in the system is conserved

$mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s$

The maximum angular velocity is 5.82812 rad/s

4 0

2 years ago

A source of sound is kept in a jar in a vacuum. Air is slowly introduced in to the jar. What happens to the sound coming out of

timama [110]

<span>The loudness of the sound increases gradually as the air is slowly introduced in to the jar. This is because sound needs a physical medium and in a vacuum there is none. The air provides that medium and as it is introduced, the transfer of sound energy increases</span>

3 0

2 years ago

. Now assuming Anna's far point was found to be 0.9 m (i.e., her eyes can't focus on any object more than 0.9 m away), what powe

Mama L [17]

Answer:1). Distance of far point x=0.9m

Therefore, since the image is virtual

-f=-x = -0.9m

Power of the concave lenses = 1/f = 1/-0.9

= -1.11D

2 ) near point is 21cm = 0.21m

Power = 4-1/near point

= 4/0.21

= 14.2D.

7 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0

2 years ago