Ask question

Ask question

All categories

3 years ago

9

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull

s with 61.7 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 147 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey. Magnitude of the net force: _____N What is the direction of the net force? Express this as the angle from straight ahead between 0° and 90°, with a positive sign for angles to the left and a negative sign for angles to the right.
Direction of the net force: _____°

1 answer:

DiKsa [7]3 years ago

3 0

Answer:

$F_{net} = 232.8 N$

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

$F_x = F_1 + F_2cos45 + F_3cos45$

here we know that

$F_1 = 77.3 N$

$F_2 = 61.7 N$

$F_3 = 147 N$

$F_x = 77.3 + 61.7 cos45 + 147 cos45$

$F_x = 224.9 N$

Similarly in Y direction we will have

$F_y = F_3 sin45 - F_2 sin45$

$F_y = (147 - 61.7)sin45$

$F_y = 60.3 N$

Now the net force on the donkey is given as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$F_{net} = \sqrt{224.9^2 + 60.3^2}$

$F_{net} = 232.8 N$

Now direction of force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{60.3}{224.9}$

$\theta = 15^o$ towards right so it is -15 degree

You might be interested in

You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is

tangare [24]

$1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }$

<u>Explanation</u>:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

$\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}$

Where, F = force, $G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right)$ , $mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}$ , $mass \left M_{2}= 0.31 \mathrm{kg}$ and Distance(r) = 0.42 m.

Substitute the values in the formula.

$\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}$

$\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}$

$\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}$

$\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }$

3 0

4 years ago

Officials begin to release water from a full man-made lake at a rate that would empty the lake in 4 weeks, but a river that can

iris [78.8K]

Answer:

5 weeks and 5 days is required to empty the lake

Explanation:

Officials begin the remove water from a full man made lake

The lake can be emptied in 4 weeks

= -1/4

A river can fill the lake up in 15 weeks

= 1/15

Let t represent the number of weeks that is required to empty the lake

= -1/t

Therefore the number of weeks it takes to empty the lake can be calculated as follows

-1/t= -1/4 + 1/15

-1/t= -11/60

Cross multiply

-11×t= -1×60

-11t= -60

t = 60/11

t= 5 5/11

Hence it takes 5 weeks and 5 days to empty the lake

6 0

4 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

3 years ago

HELP ME ASAP!!!!100 points!!!!!<br><br><br><br><br>How do you get your fingers in the Labia?

Nina [5.8K]

You literally just put your fingers in your genitals? is this for a sex ed course...?

4 0

3 years ago

Read 2 more answers

The average intensity of sunlight at the top of the earth's atmosphere in 1390 w/m2. what is the maximum energy that a 34-m x 46

slavikrds [6]

Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
7200s×2173960w =15,652,512,000 joules of energy

7 0

3 years ago