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inessss [21]
3 years ago
12

The sketch shows a painter’s scaffold in mechanical equilibrium. The person in the middle weighs 225 N, and the tension in each

rope is 300 N. What is the weight of the scaffold?
Physics
1 answer:
musickatia [10]3 years ago
3 0

Answer:

The weight of the scaffold is 375 N.

Explanation:

It is given that,

The weight of the person in the middle, W = 225 N

Tn tension in each rope, T = 300 N

Let w is the weight of the scaffold. On equilibrium, the net forces can be written in terms of :

w + 225 = 300 + 300

w = 375 N    

So, the weight of the scaffold is 375 N. Hence, this is the required solution.      

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Two factors that regulate (control) glandular secretion.
antiseptic1488 [7]

Answer:

The factors include age and puberty

Explanation:

Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.

As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.

Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.

6 0
3 years ago
Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten
ladessa [460]

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

m_1v_1 + m_2v_2 = 0

now plug in all data into above equation

5(v) + 3(3)

5v = -9

v = -1.8 m/s

so correct answer is

A) - 1.8 m/s

3 0
3 years ago
Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it
olga2289 [7]
<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>
6 0
3 years ago
A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft
erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)

Now, differentiating both sides with respect to time 't', we get:

2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}

Now, we are given speed as 25 mph. So, \frac{de}{dt}=25\ mph

Also, when e=91\ mi, we can find 'x' using equation (1). This gives,

x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi

Now, plug in the values of 'e' and 'x' and solve for \frac{dx}{dt}. This gives,

\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0
3 years ago
Cual sera el caudal que lleva un rio cuando se desplaza a 200 litros cada 40 segundos
alisha [4.7K]

Answer:

Q = 5 L/s

Explanation:

To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):

Q=\frac{V}{t}

V: Volume (volumen) = 200L

t: time (tiempo) = 40 s

you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):

Q=\frac{200L}{40s}=5\frac{L}{s}

Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)

4 0
3 years ago
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