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3 years ago

10

A block lies on a frictionless floor. a force of 5 n pulls toward the east while a force of 4 n pulls toward the north. what is

the magnitude of the net force on the block?

1 answer:

sveticcg [70]3 years ago

3 0

Given below the arrangement of loading on the larger boat by two tug boats.

F₁ = 5 N

F₂ = 4 N

Angle between them θ = 90⁰

Resultant between two vectors, $F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta }$

Substituting

$F = \sqrt{5^2+4^2+2*5*4*cos 90} \\ \\ = 6.403 N$

So magnitude of the net force on the block = 6.403 N

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The acceleration due to gravity on the moon is 1.6 m/sec. what does a 10kg mass weigh on the moon

marysya [2.9K]

Weight is a force so by F = ma, we have
weight = 10 x 1.6
= 16 N

6 0

3 years ago

A 26.4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the ca

MA_775_DIABLO [31]

Answer:

The horizontal component of the force exerted by the hinge on the beam is 47.15 N.

Explanation:

Given data:

Weight of beam = 26.4 kg

Angle between the beam and the cable is 90°

Beam inclination with respect to horizontal with an angle, $\theta = 23.4\°$

<u>We need to find the horizontal component of the force exerted by the hinge on the beam.</u>

Solution:

Let 'L' be length of the beam, 'T' be tension in the cable , $F_{h}$ be horizontal component of force by the hinge, and $F_{v}$ be vertical component of force by the hinge.

Take counterclockwise torque as positive.

Let us find torques around the hinge.

Torque by tension is given as:

$\tau = T \times L$

Torque by the force of gravity is given as:

$\tau_g= m g \frac{L}{2}\times cos \theta$

Torques by $F_{h}$ and $F_{v}$ are 0 as they act on the hinge itself.

Now, for equilibrium, net torque about the hinge is 0. So,

$\tau-\tau_g=0$

$T L - m g \frac{L}{2}\times \cos(\theta) = 0$

Dividing both sides by 'L', we get:

$T - m \frac{g}{2}\times \cos \theta = 0$

$T=m \frac{g}{2} \times cos \theta$ --------------------(1)

As per question, the cable makes 90° with the horizontal.

So, the net horizontal force is also zero. Therefore,

$F_{h} -T cos(90- \theta) = 0$

$F_h - T sin(\theta) = 0$

$F_h = T sin(\theta)$ --------------------------(2)

Plug the value of 'T' from equation (1) into equation (2). This gives,

$F_{h} = m \frac{g}{2} \times cos \theta \times sin \theta$

$F_{h} = 26.4 \times \frac{9.8}{2} \times cos(23.4) \times sin(23.4)$

$F_{h} = 47.15\ N$

Therefore, the horizontal component of the force exerted by the hinge on the beam is 47.15 N.

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3 years ago

How is pressure related to force and surface area

kherson [118]

Pressure is Force per Unit Area.Pressure is the force on an object that is spread over a surface area. The equation for pressure is the forcedivided by the area where the force is applied.

7 0

3 years ago

Two objects are attracted to each other by electromagnetic forces. Suppose I double the charge of one object. How can I return t

nata0808 [166]

Answer:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

Explanation:

Lets assume,

Charge on first object = Q

Charge on second object = q

Distance between them = r

Force between the two charges = F

According to Coulomb's law,

$F = k \frac{Qq}{r^{2}}$

where, k = Coulomb constant

New value of charge on first object = 2Q. Thus the new force(F') will be

$F' = k \frac{2Qq}{r^{2}}$

$F' = 2F$

So, to bring the value of force(F') to original value, there are two options:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

4 0

3 years ago

Hola puntos gratis !!!!!!!!!!!!!!!!!

mina [271]

Answer:

gracias

Explanation:

7 0

3 years ago

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