Ask question

Ask question

All categories

1 year ago

6

A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 N and 15

10 N, respectively. The acceleration of the cable is 0.620 m/s2, upward. What is the tension in the cable(a) below the worker and(b) above the worker?

1 answer:

Alex777 [14]1 year ago

4 0

Answer:(a)2631.46 N

(b)1605.51 N

Explanation:

Assuming there is a cable between crate and worker

Given

weight of worker $W_w=965 N$

Weight of crate $W_c=1510 N$

mass of worker $m_w=\frac{965}{g}=98.46 kg$

mass of crate $m_c=\frac{1510}{g}=154.08 kg$

acceleration of system $a=0.620 m/s^2$

Tension in the wire carrying Worker

$T-(m_w+m_c)g=(m_w+m_c)a$

$T=(m_w+m_c)(g+a)$

$T=(252.54)(9.8+0.62)$

$T=2631.46 N$

(b)Tension in the string above

$T_2-m_c\cdot=m_ca$

$T_2=m_c(g+a)$

$T_2=154.08(9.8+0.62)$

$T_2=1605.51 N$

You might be interested in

When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0

1 year ago

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

Alja [10]

Answer:

Terminal velocity of object = 12.58 m/s

Explanation:

We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = $12.0v+4.00v^2$

Equating both, we have

$784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0$

So v = 12.58 m/s or v = -15.58 m/s ( not possible)

So terminal velocity of object = 12.58 m/s

7 0

1 year ago

Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 × 109 light-y

-BARSIC- [3]

Answer:

$3.7\times 10^{51})$ kg

Explanation:

$R$ = radius of the sphere modeled as universe = $13\times 10^{25}$ m

Volume of sphere is given as

$V = \frac{4\pi R^{3}}{3}$

$V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}$

$V = 9.2\times 10^{78}$ m³

$\rho$ = average total mass density of universe = $1\times 10^{-26}$ kg/m³

$m$ = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

$m = \rho V$

$m = (1\times 10^{-26}) (9.2\times 10^{78})$

$m = 9.2\times 10^{52}$ kg

$M$ = mass of "ordinary" matter = ?

mass of "ordinary" matter is only about 4% of total mass, hence

$M = (0.04) m$

$M = (0.04)(9.2\times 10^{52})$

$M = 3.7\times 10^{51}$ kg

6 0

1 year ago

During the time a 2.0 kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m abov

Brums [2.3K]

Answer:

294J

Explanation:

The work done by gravity is independent of the horizontal distance. It is only dependent on the vertical height of 15m above its initial position.

The work done against gravity is given by

$W=mgh....................(1)$

Given;

m = 2.0kg

$g=9.8m/s^2$

Hence;

$W = 2.0*9.8*15\\W=294J$

4 0

2 years ago

A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T

irina1246 [14]

Answer:

Option d)

Solution:

As per the question:

Work done by farm hand, $W_{FH} = 972J$

Force exerted, F' = 310 N

Angle, $\theta = 23^{\circ}$

Now,

The component of force acting horizontally is F'cos $\theta$

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

$W_{FH} = \vec{F'}.\vec{d}$

$972 = 310\times dcos23^{\circ}$

d = 3.406 m = 3.4 m

3 0

2 years ago