Ask question

Ask question

All categories

3 years ago

8

Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The

rope makes an angle of 46.6 ◦ with the horizontal. If Beth exerts a force of 124 N on the rope, what is the magnitude of the upward component of the force acting on the stake? Answer in units of N.

2 answers:

valina [46]3 years ago

7 0

Answer:

90.1 N

Explanation:

We are given that

The ropes makes an angle with horizontal= $46.6^{\circ}$

Beth exerts a force on the rope=124 N

We have to magnitude of the upward component of the force on the stake.

Vertical component is always sin and horizontal component is always cos.

Therefore, the magnitude of the upward component of the force acting on the stake= $124 sin 46.6^{\circ}$

The magnitude of the upward component of the force acting on the stake= $124 \times 0.727$

The magnitude of the upward component of the force acting on the stake=90.148 N

Hence, the magnitude of the upward component of force acting on the stake=90.1 N

Delvig [45]3 years ago

6 0

Answer:

The upward component force in magnitude is 90.095 N

Solution:

As per the question:

The angle that the rope make with the horizontal, $\theta = 46.6^{\circ}$

The force exerted by Beth on the rope, F = 124 N

Now, the upward component of the force that acts on the stake is the vertical component or the cosine component of the force that Beth exerts on the rope.

The magnitude of this component of force is given by:

$F_{up} = Fsin\theta$

$F_{up} = 124\times sin46.6^{\circ} =90.095 N$

$F_{up} = 90.095 N$

You might be interested in

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius

Gennadij [26K]

Answer:

t = 0.0735 m

Explanation:

Angular acceleration of the flywheel is given as

$\alpha = 3 rad/s^2$

now after t = 8 s the speed of the flywheel is given as

$\omega = \alpha t$

$\omega = 3 \times 8$

$\omega = 24 rad/s$

now rotational kinetic energy of the wheel is given as

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(\frac{1}{2}mR^2)(24^2)$

$800 = \frac{1}{4}m(0.23)^2(24^2)$

$m = 105 kg$

now we have

$m = \rho (\pi R^2) t$

$105 = 8600(\pi \times 0.23^2) t$

$t = 0.0735 m$

4 0

3 years ago

How does a mathematical model help you understand the science concepts

Dmitrij [34]

It can help with measurements and when you want to add measurements to a cylinder or a beaker so ya

7 0

3 years ago

Which statement best describes what energy transfer diagrams show? Energy can change form, but the total amount of energy stays

Rom4ik [11]

Energy can change form, but the total amount of energy stays the same.

3 0

3 years ago

Read 2 more answers

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

$\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

${\frac{M_{B}}{M_{A}}} = 4$

${\frac{M_{B}}{4}} = M_{A}$

7 0

3 years ago