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3 years ago

3. What are the two parts of a shadow?

Physics

1 answer:

Zielflug [23.3K]3 years ago

5 0

Answer:

the umbra and penumbra

Explanation:

The two parts of a shadow are the umbra and penumbra.

A shadow is an expression formed when light from a source is cut blocked by an opaque body.

Most shadows are made up of two parts, umbra and penumbra.

The umbra is the darkest, and the innermost part of a shadow.
In the penumbra, only a portion of the light source is blocked. It is more of like a partial shadow.

The umbra is directly formed by light which impinges on an opaque body and it completely cut off.

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A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

atroni [7]

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

bullet mass, Mb = 4.10 g

initial bullet velocity, Vbi = 837 m/s

wooden block mass, Mw = 820 g

initial wooden block velocity, Vwi = 0 m/s

final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

6 0

3 years ago

Which is a product of nuclear fusion?

Tju [1.3M]

Answer:

Creation of energy by joining the nuclei of two hydrogen atoms to form helium

7 0

3 years ago

Read 2 more answers

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

3 years ago

Calculate the percent error of the distanc

Soloha48 [4]

Answer:

5.25%

Explanation:

From the question given above, the following data were obtained:

Accepted value = 238857 miles

Measured value = 226316 miles

Percentage error =.?

Next, we shall determine the absolute error. This can be obtained as follow:

Accepted value = 238857 miles

Measured value = 226316 miles

Absolute Error =?

Absolute Error = |Measured – Accepted|

Absolute Error = |226316 – 238857|

Absolute Error = 12541

Finally, we shall determine the percentage error. This can be obtained as follow:

Accepted value = 238857 miles

Absolute Error = 12541

Percentage error =.?

Percentage error = absolute error / accepted value × 100

Percentage error

= 12541 / 238857 × 100

= 1254100 / 238857

= 5.25%

Therefore, the percentage error is 5.25%.

8 0

3 years ago

Walt ran 5 km in 25 minutes going east to what was his average velocity

goldenfox [79]

distance d = 5 km = 5 x 1000 m = 5000 m

time taken = 25 minute = 25 x 60 sec = 1500 sec

average velocity V = d/t

V = 5000/1500

V = 3.33 m/s towards east

3 0

3 years ago

Read 2 more answers