Ask question

Ask question

All categories

2 years ago

12

A 67.0 kg crate is being raised by means of a rope. Its upward acceleration is 3.50 m/s2. What is the force exerted by the rope

on the crate? A 67.0 kg crate is being raised by means of a rope. Its upward acceleration is 3.50 m/s2. What is the force exerted by the rope on the crate?

1 answer:

serious [3.7K]2 years ago

7 0

<span>A 67.0 kg crate is being raised by means of a rope. Its upward acceleration is 3.50 m/s2. What is the force exerted by the rope on the crate?

</span>Newton's Second Law<span> of Motion states, “The force acting on an object is equal to the mass of that object times its acceleration.” We calculate as follows:
</span>
F = ma = 67.0 kg (3.50 m/s^2) = 234.5 J

You might be interested in

2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA

11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *144

M_c = 2.5*( 42 / 150 ) *144

M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

5 0

3 years ago

Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu

Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density $D$ appears as a physical characteristic property of matter that establishes a relationship between the mass $m$ of a body or substance and the volume $V$ it occupies:

$D=\frac{m}{V}$ (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is $m=0.155g$ and th volume $V=0.000235L=0.235mL$

Solving (1) with these values:

$D=\frac{0.155g}{0.235mL}$ (2)

$D=0.660g/mL$ (3)

<h2>Answer B:</h2>

In this case the mass of a sample is $m=4.71g$ and its density is $D=3.63g/mL$ .

Isolating $V$ from (1):

$V=\frac{m}{D}$ (4)

$V=\frac{4.71g}{3.63g/mL}$ (5)

$V=1.297mL$ (5)

<h2>Answer C:</h2>

In this case the volume of a sample is $V=0.293mL$ and its density is $D=0.930g/mL$ .

Isolating $m$ from (1):

$m=D.V$ (6)

$m=(0.930g/mL)(0.293mL)$ (7)

$m=0.272g$ (8)

4 0

3 years ago

Read 2 more answers

How much support force does a table exert on a book that weighs 15 N when the book is placed on the table?

mestny [16]

Answer:

15 N

Explanation:

According to Newton's third law of motion, to every action, there is an equal and opposite reaction. This reaction is equal in magnitude to the force acting but in an opposite direction.

Now, if the book weighs 15 N, an opposite equal force will be: N = -15 N

But the magnitude of this will be the absolute value which is 15N.

8 0

2 years ago

If you have two uncertainties, and they are from two different sources and contribute to the uncertainty of a measurement, what

Darya [45]

The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.

Δm = ∑ $| \frac{dm}{dx_i} | \ \Delta x_i$

Physical quantities are precise values of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the cases, all the errors add up.

If m is the calculated quantity, x_i the measured values and Δx_i the uncertainty of each value, the total uncertainty is

Δm = ∑ $| \frac{dm}{dx_i } | \ \Delta x_i$ | dm / dx_i | Dx_i

for instance:

If the magnitude is a average of two magnitudes measured each with a different error

m = $\frac{m_1+m_2}{2}$

Δm = | $\frac{dm}{dx_1}$ | Δx₁ + | $\frac{dm}{dx_2}$ | Δx₂

$\frac{dm}{dx_1}$ = ½

$\frac{dm}{dx_2}$ = ½

Δm = $\frac{1}{2}$ Δx₁ + ½ Δx₂

Δm = Δx₁ + Δx₂

In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.

Learn more about propagation errors here:

brainly.com/question/17175455

6 0

2 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2010 × 103 seconds (about

vazorg [7]

Answer:

Radial acceleration of moon is $a_{r} = 2.246\times 10^{-3}$ $\frac{m}{s^{2} }$

Explanation:

Given :

Time period $T = 1.987 \times 10^{6}$ sec

Distance from center of moon to planet $r = 225 \times 10^{6}$ m

From the equation of radial acceleration,

$a_{r} = r\omega ^{2}$

Where $\omega = 2\pi f = \frac{2\pi }{T}$

So $\omega = 3.16 \times 10^{-6} \frac{rad}{s}$

Now moon's radial acceleration,

$a_{r} = 225 \times 10^{6} \times (3.16 \times 10^{-6} )^{2}$

$a_{r} = 2246.76 \times 10^{-6}$

$a_{r} = 2.246\times 10^{-3}$ $\frac{m}{s^{2} }$

7 0

3 years ago