Ask question

Ask question

All categories

3 years ago

7

A 50 kg rock climber accidentally falls from the side of a rock and freefalls until she is stopped by her 9 m saftey rope. Assum

ing that the rope stops her completely with no rebound, what is the impulse impared on her body by the rope? If the rock climber's rope took 0.224s to stop her completely after she fell, what was the average force felt by the rock climber as she stopped?

1 answer:

olchik [2.2K]3 years ago

8 0

Answer:

a) 665 kg.m/s

b) 2.97 kN

Explanation:

The impulse is given by:

$J=\Delta p\\J=m(v_a-v_b)$

where

va=velocity after

vb=velocity before

The velocity just before is given by:

$v_b=\sqrt{2*9.81m/s^2*9m}=13.3m/s(-\hat{j})$

The velocity just after is zero because there wasn't a rebound.

So the impulse is:

$J=50kg*(0-13.3m/s(-\hat{j}))\\J=50kg*(0+13.3m/s(\hat{j}))=665kg.m/s$

The impulse is also given by:

$J=\Delta p=F* \Delta t\\so:\\F=\frac{J}{\Delta t}\\\\F=\frac{665kg.m/s}{0.224s}=2.97kN$

You might be interested in

How much is the tension number 2

mestny [16]

This is what I got:

Net force in the Y direction:

ΣFy = T1 - T2

F = ma

ma = T1 - T2

Isolate for T2

ma - T1 = -T2

Multiply by -1

T1 - ma = T2

100 - (3)(2) = T2

100 - 6 = T2

T2 = 94 N

4 0

3 years ago

What is the unit for work is called

emmasim [6.3K]

Answer:

The unit you should use for work done and energy is the joule (J) which is indeed the same as the newton metre (N m).

There is another physical quantity which is the product of force and distance and that is torque or moment of a force.

The unit you should use for torque is the newton metre (Nm) and not the joule.

Naming the units of work done and torque differently helps to emphasis the fact that work done and torque refer to two different physical quantities although the definitions of both quantities have the product of force and distance in them.

work done=force→⋅displacement→ and torque→=force→×displacement→

Hope I helped

4 0

4 years ago

What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol

san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

$Q=\frac{\Delta H}{n}$

where,

Q = energy absorbed = 27.2 kJ

$\Delta H$ = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

$27.2kJ=\frac{6.01kJ/mol}{n}$

$n=0.221mol$

Now we have to calculate the mass of ice.

$\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}$

Molar mass of ice = 18.02 g/mol

$\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g$

Thus, the mass of ice melted can be, 3.98 grams.

3 0

4 years ago

Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 494 Hz when air temperature i

elena-14-01-66 [18.8K]

Answer:

0.173 m.

Explanation:

The fundamental frequency of a closed pipe is given as

fc = v/4l .................. Equation 1

Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.

Making l the subject of the equation,

l = v/4fc ................ Equation 2

also

v = 331.5×0.6T ................. Equation 3

Where T = temperature in °C, T = 18.0 °c

Substitute into equation 3

v = 331.5+0.6(18)

v = 331.5+10.8

v = 342.3 m/s.

Also given: fc = 494 Hz,

Substitute into equation 2

l = 342.3/(4×494)

l = 342.3/1976

l =0.173 m.

Hence the length of the organ pipe = 0.173 m.

7 0

3 years ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0

3 years ago