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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.
Physics
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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Answer:

Answer; v= 1.2654m/s

T= 110.76N

Explanation:

Apply Momentum Principle

Fdtro - Mgridt = Iow +Mvr

Fdtro - Mgridt = mK2 v/r1 + Mvr1

85 x 3x 0.345 -11 x 9.81 x 0.23 x 3 =30 x 0.25 x 0.25 x v/0.23 + 11 x v x 0.23 =

v = 1.2654m/s

To find the timed average value

Tdt -Mgdt =MV

T x 3 - 11 x 9.81 x 3 = 11 x 0.778

T= 110.76N

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2 years ago
Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th
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Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0
1 year ago
Read 2 more answers
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
Makovka662 [10]

Answer:

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k

a=5 i+2t j - 6\ t^2k

Therefore the acceleration function a will be

a=5 i+2t j - 6\ t^2k

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

a=\sqrt{5^2+4^2+24^2}\ m/s^2

a= 24.83 m/s²

The direction of the acceleration a is given as

a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

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2 years ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

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Both hit the ground at the same time

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h=\frac{gt^2}{2}

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13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

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1 year ago
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Answer:

yes independent of the sign or valve of Q

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