"NEED HELP FAST" Which of the following is a vector quantity?

1) A person having a mass of 59.1 kg stands before a flight of 30 stairs each of which is 25.0 cm high. He runs up 20 stairs, tu

poizon [28]

Answer:

P = 147,75 W

Explanation:

A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high

Height at 30 steps , h=30×2.5= 7.5 m

Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J

So, Work done by the man , W= 4432.5J

Power used , P= $\frac{W}{T}$

P = 4432.5 /30

P = 147,75 W

Solve any question of Work, Energy and Power with:-

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3 0

1 year ago

Question 9

egoroff_w [7]

Answer:

C

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a

=36÷2

= 18

6 0

2 years ago

4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.

kolbaska11 [484]

Answer:

$\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}$

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

$\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg$

6 0

3 years ago

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = $\frac{1}{4}$ F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.

6 0

3 years ago

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

2 years ago