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Yakvenalex [24]

2 years ago

5

Part E Two forces, of magnitude 4N4N and 10N10N, are applied to an object. The relative direction of the forces is unknown. The

net force acting on the object __________. Check all that apply.

1 answer:

Sedaia [141]2 years ago

6 0

Answer:

A.) cannot be equal to 5N

Explanation:

Forces are vectors: therefore, when two forces act together on the same object, in order to find the resultant force we must apply the rule of additions for vectors.

Therefore, given two forces $F_1,F_2$ , the magnitude of the resultant force can have:

- A minimum value of $|F_1-F_2|$ , if the two forces have opposite direction

- A maximum value of $|F_1+F_2|$ , if the two forces have same direction

In this problem, the two forces that we have are

$F_1=4 N\\F_2=10 N$

Therefore, the minimum and maximum magnitude of the net force is

$F_{min}=|4-10|=6 N\\F_{max}=|4+10|=14 N$

So the correct statement is

A.) cannot be equal to 5N

Other statements are wrong because:

B.) cannot be equal to 10N --> 10 N is between the interval 6-14 N

C.) cannot be directed the same way as the force of 10N --> the net force has the same direction of the 10 N force, if also the 4N force has the same direction

D.) must be greater than 10N --> the net force can have up to a magnitude of 14 N

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To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.

lozanna [386]

Answer:

a)

Explanation:

A block sliding down an inclined plane, is subject to two external forces along the slide.
One is the component of gravity (the weight) parallel to the incline.
If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

$F_{gp} = m*g* sin \theta (1)$

(taking as positive the direction of the movement of the block)

The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

$F_{f} = \mu_{k} * N (2)$

The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

$N = m*g* cos \theta (3)$

In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

$m*g* sin \theta = \mu_{k} * m*g* cos \theta$

As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

$\mu_{k} = tg \theta$

8 0

1 year ago

A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes t

BARSIC [14]

Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

Suppose $F_a$ and $F_b$ be the average force exerted on A and B respectively

and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.

Thus $F_a=-F_b$ and option d is correct

4 0

2 years ago

A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi

tekilochka [14]

This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N

The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
= 1721.63 N</span>

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely, so we can say that the horizontal component is zero:

Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
= 811.66 N</span>

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0

2 years ago

Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1m x 1m x 0.10m slab of foam, so

goblinko [34]

Answer: Weight = 98.1N

Explanation:

Density of water = 1000 kg/m^3

Given that the Plastic foam is about 0.10 times as dense as water. That is,

Density of plastic foam = 0.1 × 1000 = 100kg/m^3

The volume V = 1 ×1×0.1 = 0.1 m^3

Density is the ratio of mass to volume

Density = mass/volume

Let us substitute for density and volume to get mass.

100 = M/0.1

Make M the subject of formula

M = 100 × 0.1 = 10 kg

Weight = mg

Where g = 9.81 m/s

Substitute the M and g into the formula

Weight = 10 × 9.81 = 98.1 N

Therefore, the weight of the brick is 98.1 N

4 0

2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

$I=\frac{Q}{t}$

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

$I=\frac{420 C}{4800 s}=0.0875 A$

2) Drift velocity of the electrons: $1.78\cdot 10^{-6} m/s$

The drift velocity of the electrons in the wire is given by:

$u = \frac{I}{nAq}$

where

I = 0.0875 A is the current

$n=5.8\cdot 10^{28}$ is the number of free electrons per cubic meter

A is the cross-sectional area

$q=1.6\cdot 10^{-19} C$ is the charge of one electron

The radius of the wire is

$r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m$

So the cross-sectional area is

$A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2$

So, the drift velocity is

$u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s$

4 0

2 years ago