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3 years ago

Please I need help with this question.

Physics

1 answer:

gavmur [86]3 years ago

5 0

Answer:

m1 = √2kg , m2 = 1kg , m3 = 1kg

Explanation:

Since the mass m1 can't move, the sum of horizontal and vertical forces must be zero. Since the mass m1 is suspended symmetrically the horizontal forces are equal if mass m2 and m3 are equal.

The tension in each string suspending mass m1 must match the force of gravity pulling on m1. The tension in each string is:

T = m2*g = m3*g = g

The vertical forces pulling m1 up is therefore: 2 * T * cos 45° = √2 * T = √2 * g

This force must match the force of gravity G pulling m1 down. G = m1 * g

Combining both equations: √2 * g = m1 * g

m1 = √2

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The answer to this is c. High pressure causes plastic flow to occur.

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3 years ago

Children in a tree house lift a small dog in a basket 4.90m up to their house. If it takes 201J of work to do this, what is the

scZoUnD [109]

Answer:

4.18 kg.

Explanation:

Work done: Work is said to be done when ever a force move a body through a given distance. The S.I unit of work done is Joules (J)

Mathematically, work done is expressed as

W' = F×d.................... Equation 1

Where W' = work done, F = force , d = distance.

making F the subject of the equation,

F = W'/d............................. Equation 2

Note: The force need to lift the small dog n a basket = combined weight of the dog ans the basket.

Therefore,

W = F

Where W = combined weight of the dog and the basket.

Also

W = Mg

M = W/g............................. Equation 3

Where M = combined mass of the dog and the basket, g = acceleration due to gravity.

Given: W' = 201 J, d = 4.90 m.

Substitute into equation 2

F = 201/4.9

F = 41.02 N.

Since F = W = 41.02 N and g = 9.81 m/s²

Substitute these values into equation 3

M = 41.02/9.81

M = 4.18 kg.

Thus the combined mass of the dog and the basket = 4.18 kg.

4 0

3 years ago

Nathan has a car that accelerates at 5 m/s2. If the car has a mass of

zhannawk [14.2K]

Answer:

F = 5000 [N] or 5 [kN]

Explanation:

In order to solve this problem we must take into account Newton's second law, which tells us that the sum of all forces applied on a body must be equal to the product of the mass of the body by its acceleration.

F = m*a

where:

F = force [N]

m = mass = 1000 [kg]

a = acceleration = 5 [m/s^2]

F = (1000*5)

F = 5000 [N] or 5 [kN]

5 0

3 years ago

. In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscilla

cluponka [151]

Answer:

Spring constant, k = 1.16 N/m

Explanation:

It is given that,

Mass of the air track, m = 0.2 kg

Time, t = 2.6 s

Let T is the time period of the spring. The expression for the time and spring constant is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.2}{(2.6)^2}$

k = 1.16 N/m

So, the spring’s force constant is 1.16 N/m. Hence, this is the required solution.

6 0

3 years ago

The deepest point of the pacific ocean is 11,033 m, in the mariana trench. what is the gauge pressure in the water at that point

Liula [17]

Given that,

Depth of seawater, h = 11,033 m

Density seawater, p (rho) = 1025 kg/m³

Gauge Pressure , P = ??

Since, we know that:

Pressure, P = pgh

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Pressure = 1109395723.3 N/m²

Pressure = 1.1 x 10∧8 Pascal

4 0

3 years ago