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Paladinen [302]
2 years ago
6

A man walks 2.6 miles east, then turns and walks 6.7 miles north. What is his resultant

Physics
1 answer:
Zarrin [17]2 years ago
6 0

Answer:

Yes cause he walks 6.7 miles

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How do you know if a boy like you?
adell [148]

Answer:

They cherish you

Explanation:

6 0
3 years ago
U = 3, 9 , v = 4, 2 (a) find the projection of u onto v. (b) find the vector component of u orthogonal to v.
ExtremeBDS [4]

Answer:

(a) At U = 3, 9 , v = 4, 2, the projection of u onto v is w1=<2,8>

(b)At U = 3, 9 , v = 4, 2,  the vector component of u orthogonal to v is w2 = <4,-1>

Explanation:

A

The projection of u onto v is given by:

w1= projvu = (u⋅v||v||2)v

Given that  u= <6,7> and v=<1,4>, we can find the projection of u onto v as shown below:

w1= projvu = (u⋅v||v||2v=(<6,7>⋅<1,4><1,4>⋅<1,)

=(6⋅1+7⋅41⋅1+4⋅4)<1,4>

=3417<1,4>

=<2,8>

Part B

The vector component of u orthogonal to v is given by:

Using the given vectors and the projection found in part (a), we can find the vector component of u orthogonal to v as shown below:

w2=u−projvu

=<6,7≻<2,8>

=<(6−2),(7−8)>

=<4,−1>

To learn more about vector component, click brainly.com/question/17016695

#SPJ4

5 0
1 year ago
0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compre
Dmitry_Shevchenko [17]

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

so

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e²  = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

5 0
3 years ago
A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me
loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

Where,

m = mass

C_p= Specific Heat

\Delta T = Change in temperature

Replacing with our values we have that

m = 0.1g

C_p = 139J/Kg\cdot K \rightarrow Specific heat of mercury

\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K

Replacing

Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J

Therefore the heat lost by mercury is 0.09J

5 0
3 years ago
A cannon launches a 4.0kg bowling ball with 50J of kinetic energy what is the bowling ball’s speed
kobusy [5.1K]

Explanation:

I will try to find the formula

4 0
3 years ago
Read 2 more answers
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