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11111nata11111 [884]
3 years ago
5

Every winter I fly to Michigan. It takes 5 hours. What is my average speed?

Physics
1 answer:
VladimirAG [237]3 years ago
6 0
Your answer is 780 km/hr
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Ramas weight is 40 kg. She is carrying a load of 20 kg up to a height of 20 m. What work does she do? What work does she do?
Taya2010 [7]

Answer:

The work done by Ramas is 3920 N.

Explanation:

Given;

mass of the load lifted by Ramas, m = 20 kg

height through which the load is lifted, h = 20 m

The work done by Ramas is equal to gravitational potential due to the height in which the load is lifted.

W = PE = mgh

where;

g is acceleration due to gravity = 9.8 m/s²

W = 20 x 9.8 x 20

W = 3920 N.

Therefore, the work done by Ramas is 3920 N.

3 0
3 years ago
The Clean Air Act emphasizes that one way to prevent and reduce air pollution is to involve public participation
Marina CMI [18]
I think the answer would be true that’s my opinion
7 0
4 years ago
10 moles of sulfur dioxide gas (Pitzer accentric factor = 0.25, critical temperature = 430K, critical pressure = 78 atm) are pla
den301095 [7]

Answer:

vbmdgj

Explanation:

4 0
3 years ago
A penguin slides at a constant velocity of 1.43 m/s down an icy incline. The incline slopes above the horizontal at an angle of
andrew11 [14]

Answer:

t =1.285  s

Explanation:

u=1.43 m/s

∅=6.47°

We know that  μ(k)=tan∅

F(net)=-f(k)

ma =μ(k)*m*g

a =μ(k)*g   equation 1

v=u+at

0=u+(μ(k)*g*t

Putting values

t=u/(g*tan∅)

t=\frac{1.43}{9.81*tan(6.47)}

t=1.285 s

7 0
3 years ago
A meteoroid passes through a position in space where its speed is very small relative to Earth's and it is at a perpendicular di
Oliga [24]

Answer:

v=7506.4m/s

Explanation:

If we call 1 the position in space where the meteoroid of mass m speed is very small relative to Earth's (whose mass is M_E=5.98\times10^{24}kg and radius is R_E=6371000m) and it is at a perpendicular distance of h_1=19R_E above Earth's surface, and 2 the position when it is h_2=R_E above the ground, then, since in this case mechanical energy is conserved, we can write:

E_1=E_2

which means:

K_1+U_1=K_2+U_2

where K is the kinetic energy and U the gravitational potential energy. Since K_1=0J we can write:

\frac{-GM_Em}{r_1}=\frac{mv_2^2}{2}+\frac{-GM_Em}{r_2}

which means:

v_2=\sqrt{2GM_E(\frac{1}{r_2}-\frac{1}{r_1})}

And since r=R_E+h (the distance of an object to the center of the Earth is Earth's radius plus the height of the object), we have:

v_2=\sqrt{2GM_E(\frac{1}{R_E+h_2}-\frac{1}{R_E+h_1})}=\sqrt{2GM_E(\frac{1}{2R_E}-\frac{1}{20R_E})}=\sqrt{\frac{GM_E}{R_E}(1-\frac{1}{10})}=\sqrt{\frac{0.9GM_E}{R_E}}

This for our values is:

v_2=\sqrt{\frac{0.9GM_E}{R_E}}=\sqrt{\frac{0.9(6.67\times10^{-11}Nm^2/kg^2)(5.98\times10^{24}kg)}{6371000m}}=7506.4m/s

3 0
3 years ago
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