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1 year ago

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If the 5-N force and the 12-N force form a 90 degree angle, what is the magnitude of the force acting in the direction of the da

shed arrow

1 answer:

Leni [432]1 year ago

3 0

Answer:

<h2>13N</h2>

Explanation:

<em>Kindly see attached file for your reference</em>

Step one:

given data

the horizontal component of the force= 12N

the vertical component of the force= 5N

The dashed arrow represents the hypotenuse of the triangle, hence the resultant of the force system.

By implication of this, we will use the Pythagoras theorem to solve for the resultant force

Step two:

$F_R=\sqrt{F_H^2+F_V^2}\\\\F_R= \sqrt{12^2+5^2}\\\\F_R=\sqrt{144+25}\\\\F_R=\sqrt{169}\\\\F_R=13N$

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erastova [34]

Answer:

4m/s2

Explanation:

The following data were obtained from the question:

U (initial velocity) = 10m/s

V (final velocity) = 30m/s

t (time) = 5secs

a (acceleration) =?

Acceleration is the rate of change of velocity with time. It is represented mathematically as:

a = (V - U)/t

Now, with this equation i.e

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2 years ago

Read 2 more answers

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

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1 year ago

The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the

Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

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Where,

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Re-arrange to find I,

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Therefore the current density would be

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Comparing the two values we can see that the 2mm wire has a higher current density.

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1 year ago

The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?

Aloiza [94]

Answer:

The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

Explanation:

Given that,

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2 years ago

A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar

Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s

Explanation:

It is given that,

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Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

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Final momentum of the ball, $p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s$

Change in momentum of the ball, $\Delta p=p_f-p_i$

$\Delta p=12-(-12)=24\ kg-m/s$

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0

2 years ago