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2 years ago

An 850 kg elevator is accelerating upwards at 3.0 m/s2. Calculate the tension in the cable.

Physics

1 answer:

sveta [45]2 years ago

7 0

Answer:

10880N

Explanation:

We are given that

Mass of elevator, m=850 kg

Acceleration, $a=3m/s^2$

$g=9.8m/s^2$

We have to find the tension in the cable.

We know that

When elevator is accelerating upward then tension in the cable

$T=m(a+g)$

Using the formula

$T=850(3+9.8)$

$T=10880N$

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

elena-14-01-66 [18.8K]

The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

c is speed of light
E₀ is the peak electric field
B₀ is the peak magnetic field

B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

8 0

2 years ago

If one object is 103 km away and a second object is 106 km away, one could say that the second object is _____ times further awa

vaieri [72.5K]

Answer:

$1.03$

Explanation:

$\frac{object_{second}}{object_{first}} = \frac{106}{103} = 1.02912621359$

Round to three significant digits

$1.03$

7 0

3 years ago

Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro

goldenfox [79]

Answer:

tension in rope = 25.0 N

Explanation:

Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.

Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

∑F = F - W = 0

F = W

so tension in rope = F = T = 25 N

8 0

2 years ago

a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame

natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

$\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}$

where;

L = length of the bed

$\mu$ = viscosity

U = superficial velocity

$\epsilon$ = void fraction

dp = equivalent spherical diameter of bed material (m)

$\rho$ = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6 -------------- equation (1)

24A + 576B = 24.1 --------------- equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

B = 0.017014

From equation (1)

12A + 144B = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

$A = \frac{9.6 -144(0.017014}{12}$

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0

3 years ago

What is the average speed in 4 seconds going 8 meters

gulaghasi [49]

Answer:

Explanation:

speed is define as rate of change of distance or displacement

v=s/t

s=8 m

t=4s

v=8/4

v=2 m/s

6 0

3 years ago

Read 2 more answers