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RideAnS [48]
2 years ago
14

A 60kg man jumps off a 30 foot building. Assuming he falls from rest and there is negligible air resistance, how long will it ta

ke him to reach the ground?​
Physics
1 answer:
Dmitriy789 [7]2 years ago
5 0

Answer:

1.37 seconds before.... splat !

Explanation:

y = y0 + v0 t + 1/2 a t^2      y0 = 30       y = 0

                    v0 = 0 ( from rest)

                            a = - 32.2 ft/s^2

0 = 30 - 1/2 * 32.2 * t^2

t=1.366 seconds                   <u>  Note that   mass   is irrelevant.</u>

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The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built?
Sonja [21]
Neglecting friction and air resistance, the first hill must be built 4 times higher than it is now.
3 0
3 years ago
Which equation correctly relates mechanical energy, thermal energy and total of energy when there is friction present in the sys
agasfer [191]

Answer:

E total = ME + E thermal

Explanation:

APEX

4 0
3 years ago
The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids
BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0
3 years ago
Select the correct answer.
Simora [160]

Answer:

A. 2.36 Newtons

Explanation:

F = GmM/d²

F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²

F = 2.36121...

Very poor question design.

  mass of box... 1 significant digit

        distance... 2 significant digits

mass of earth... 3 significant digits

     value of G... 4 significant digits

Answer precision to 3 significant digits is not justifiable

7 0
3 years ago
Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta
tester [92]

Radius of nuclei is given by formula

R = R_oA^{1/3}

now we can say volume of the nuclei is given as

V = \frac{4}{3}\pi R_o^3* A

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

m = z*m_p + (A- z)*m_n

m_p = m_n = 1.008u

m = A*1.008u

Now density is given as

\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}

here we know that

R_0 = 1.2 fm

\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}

\rho = 2.31 * 10^{17} kg/m^3

So from above we can say that density of all nuclei is almost same.

5 0
3 years ago
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