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Nadya [2.5K]
4 years ago
11

Objects that reflect most wavelengths of visible light appear to be _________. A. clear B. black C. white D. colorful

Physics
1 answer:
Vika [28.1K]4 years ago
6 0
<span>Objects that reflect most wavelengths of visible light appear to be white.</span>
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The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0
3 years ago
Is gasoline a compound or a mixture ?
valina [46]
I believe its a mixture

3 0
3 years ago
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Defined the work of vector​
cricket20 [7]

Answer:

In physics, work is the amount of energy required to perform a given task (such as moving an object from one point to another). We start by defining the scalar product of two vectors, which is an integral part of the definition of work, and then turn to defining and using the concept of work to solve problems.

7 0
3 years ago
34000×0.4<br> ----------------<br> 0.02×200<br> In scientific notation
VashaNatasha [74]

Answer:

the simplified expression is written as 3.4 x 10³

Explanation:

Given expression;

\frac{34000\times 0.4}{0.02 \times 200}

in scientific notation, the expression is simplified as;

\frac{34000\times 0.4}{0.02 \times 200}  = \frac{13600}{4} = 3400 = 3.4 \times 10^3

Therefore, in scientific notation, the simplified expression is written as 3.4 x 10³

4 0
3 years ago
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a
gregori [183]

Answer:

Explanation:

Given:

volume of urine discharged, V=400~mL=0.4~L=4\times 10^{-4}~m^3

time taken for the discharge, t=30~s

diameter of cylindrical urethra, d=4\times10^{-3}~m

length of cylindrical urethra, l=0.2~m

density of urine, \rho=1000~kg/m^3

a)

we have volume flow rate Q:

Q=A.v & Q=\frac{V}{t}

where:

A= cross-sectional area of urethra

v= velocity of flow

A.v=\frac{V}{t}

\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}

v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}

v=1.06~m/s

b)

The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:

P=\rho.g.l

P=1000\times 9.8\times 0.2

P=1960~Pa

5 0
3 years ago
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