The difference between a need and a want gets blurry for some people because of:

A 1090 kg car moving +11.0 m/s

stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given:

$m_{1}$ = 1090 kg

$u_{1}$ = 11 m/s

$u_{2}$ = 0

v = 4.75 m/s

We need to find $m_{2}$

When substituting the given values in the above equation, we get

$(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75$

$11990=5177.5+4.75 m_{2}$

$4.75 m_{2}=11990-5177.5$

$4.75 m_{2}=6812.5$

$m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}$

6 0

2 years ago

How many electrons are in the outer energy level of an atom of carbon

xz_007 [3.2K]

Answer:

4 electrons

Explanation:

Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.

8 0

3 years ago

a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u

n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

$a_{c}$ = $\frac{v^{2} }{r}$

$a_{c}$ = 23.04/10.3

$a_{c}$ = 2.23 m/s²

It continues to fly but now with some tangential acceleration

$a_{t}$ = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

$a_{net}$ = $\sqrt{a_{c} ^{2} +a_{t} ^{2} }$

$a_{net}$ = $\sqrt{4.97 + 0.396}$

$a_{net}$ = 2.316 m/s²

So the magnitude of net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0

1 year ago

The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G

Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

7 0

2 years ago

What is it when you divide the total distance traveled by the total time

alexdok [17]

This is called average speed.Its really simple

4 0

3 years ago

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