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Lapatulllka [165]
3 years ago
11

Easy question lol

Physics
2 answers:
inna [77]3 years ago
3 0

Answer:

c. blue star

Explanation:

Oduvanchick [21]3 years ago
3 0
C. Blue Star
Got it right on Trivia Crack lol
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Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k
boyakko [2]

Answer:

parametric representation: x = u, y = v - u ,  z = - v

Explanation:

Given vectors :

i - j ,  j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + <em>u</em>a + vb

where:  r₀ = position vector

            u and v = real numbers

             a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v,  z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u,  y₀ - u + v,   z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

∴ parametric representation: x = u, y = v - u ,  z = - v

3 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
3 years ago
Can someone help me figure out how to find tension in a rope? The question is:
Simora [160]
You are welcome.......

8 0
3 years ago
Convert 10 kilowatt to watt​
Nataly [62]

Answer:

0.01 kilowatts cubed

Explanation:

just ask alexa lol

3 0
3 years ago
Read 2 more answers
A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration
Tpy6a [65]

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

v^2 = u^2 + 2as

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 m/s^2

s = 91 m

Therefore:

v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s

The velocity of the boat when it reaches the buoy is 7.5 m/s.

6 0
3 years ago
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