How much force is needed to accelerate a 66 kg skier at 2 m/sec^2?

Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his

just olya [345]

The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.

4 0

3 years ago

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The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer: A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life = 5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 = 5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0

3 years ago

The radius of Saturn is about 10 times the radius of Venus and the mass is about 100 times that of Venus. How much larger is the

lora16 [44]

let the mass of Venus is M then mass of Saturn is 100 M

similarly if the radius of Venus is R then the radius of Saturn is 10 R

now the force of gravity on a man of mass "m" at the surface of Venus is given by

$F_1 = \frac{GMm}{R^2}$

now similarly the gravitational force on the man if he is at the surface of Saturn

$F_2 = \frac{G*100M*m}{(10R)^2}$

$F_2 = \frac{GMm}{R^2}$

so here if we divide the two forces

$\frac{F_1}{F_2} = 1$

so here we can say

F1 = F2

so on both planets the gravitational force will be same

7 0

3 years ago

An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val

Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 $m/s^2$

as per the question the acceleration due to gravity is found to be 9.42 $m/s^2$ in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is - $\frac{0.381}{9.801} *100$

=3.88735 percent

the error is not so high,so it can be accepted.

now we have to know why this occurs-the equation of time period of the simple pendulum is give as- $T=2\pi\sqrt[2]{l/g}$

$g=4\pi^2\frac{l}{T^2}$

As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801 $m/s^2$

5 0

3 years ago

A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is

hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0

3 years ago