In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

3 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

3 years ago

Read 2 more answers

As stars age and become larger they become ________ and ________

sleet_krkn [62]

They become old and explode.

5 0

3 years ago

Test populations are studied. Population one is found to obey the differential equation dy1/da=o.2y1 and the population two obey

oksano4ka [1.4K]

Answer:

Population 1 indicates growth while Population 2 indicates a declining population

Explanation:

Here, using the given rate of change of the population, we want to determine which of the two is growing and which is declining

From the rate of change of both, we can determine this. Looking at the differential equation for the first one, we can see that it is of positive value. Looking at the differential equation for the second one. we can see it is of negative value

While a positive change rate indicates growth, a negative change rate will indicate otherwise

Hence, we can conclude that the one with a negative rate change will indicate a declining population

8 0

3 years ago

Briefly describe the two different categories of mechanics:

Stells [14]

Answer:

Statics which is when dealing with forces acting out on body and the possible motives of the body system. Kinetics which explains the motion that can occur in any type of situation.

Explanation:

5 0

3 years ago