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Alekssandra [29.7K]

3 years ago

13

An object O is acted upon by a 6 N force and a 5 N force. The angle between these two forces is 60°. Draw a scale diagram showin

g the forces acting on object O and determine the resultant force.

1 answer:

FrozenT [24]3 years ago

6 0

Answer:

Resultant force = $\sqrt{91}$ N

Explanation:

Rf = $\sqrt{f1²+f2²+2.f1.f2.cos60}$

Rf = $\sqrt{6²+5²+2.6.5.1/2}$

Rf = $\sqrt{36+25+30}$

Rf = $\sqrt{91}$

doesn't look right though...

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My name is Ann [436]

Answer: at night when the full moon is there the tide begans to appear higher

Explanation:

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3 years ago

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3 years ago

A particle of mass m = 13 kg moves in space under the action of a conservative force. Its potential energy is given by PE = 2xyz

Helen [10]

Answer:

$F_{x}$ = -12 N , $F_{y}$ = -80 N and $F_{z}$ = - 44 N

Explanation:

The force is related to the potential energy by the formula

F = -Δ U = - ( $\frac{dU}{dx}$ i ^ + $\frac{dU}{dy}$ j ^ + $\frac{dU}{dz}$ k ^)

It indicates the potential energy

U = 2xyz + 3z² + 4yx + 16

To solve this problem let's make the derivatives, to find each component of the force

$\frac{dU}{dx}$ = 2yz + 0 + 4y + 0 = 2yz + 4y

$\frac{dU}{dx}$ = 2xz + 0 + 4x + 0 = 2xz + 4x

$\frac{dU}{dx}$ = 2xy + 3 2z + 0 +0 = 2xy + 6z

We look for the expression for the force in each axis

$F_{x}$ = - 2yz - 4y

$F_{y}$ = -2xz -4x

$F_{z}$ = -2xy -6z

We calculate at the point P = (20 i ^ + 1j ^ + 4 k ^) m

$F_{x}$ = - 2 1 4 - 4 1

$F_{x}$ = -12 N

$F_{y}$ = - 2 20 4 - 4 20

$F_{y}$ = -80 N

$F_{z}$ = - 20 1 - 6 4

$F_{z}$ = - 44 N

We put together the expression for strength

F = (-12 i ^ - 80j ^ -44k ^) N

5 0

4 years ago

I will gib brainlyest or whatever.

astraxan [27]

Answer:

Range of the projectile: approximately $1.06 \times 10^{3}\; {\rm m}$ .

Maximum height of the projectile: approximately $80\; {\rm m}$ (approximately $45.0\; {\rm m}$ above the top of the cliff.)

The projectile was in the air for approximately $7.07\; {\rm s}$ .

The speed of the projectile would be approximately $155\; {\rm m \cdot s^{-1}}$ right before landing.

(Assumptions: drag is negligible, and that $g = 9.81\; {\rm m\cdot s^{-1}}$ .)

Explanation:

If drag is negligible, the vertical acceleration of this projectile will be constantly $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$ . The SUVAT equations will apply.

Let $\theta$ denote the initial angle of elevation of this projectile.

Initial velocity of the projectile:

vertical component: $u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}$
horizontal component: $u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}$ .

Final vertical displacement of the projectile: $x_{y} = (-35)\; {\rm m}$ (the projectile landed $35\: {\rm m}$ below the top of the cliff.)

Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the final vertical velocity $v_{y}$ of this projectile:

${v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}$ .

$\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}$ .

(Negative since the projectile will be travelling downward towards the ground.)

Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: $v_{x} = u_{x}$ .

The overall final velocity of this projectile will be:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}$ .

Change in the vertical component of the velocity of this projectile:

$\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}$ .

Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:

$\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}$ .

Hence, the projectile would be in the air for approximately $7.07\; {\rm s}$ .

Also the horizontal velocity of this projectile is $u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}$ throughout the flight, the range of this projectile will be:

$\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}$ .

When this projectile is at maximum height, its vertical velocity will be $0$ . Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the maximum height of the projectile (relative to the top of the $35\; {\rm m}$ cliff.)

$\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}$ .

Thus, the maximum height of the projectile relative to the ground will be approximately $45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}$ .

5 0

1 year ago

Everyday lifestyle choices affect your physical fitness. Please select the best answer from the choices provided. T F

amid [387]

Answer:
True

Explanation:
Physical fitness is defined as the health state of the body and its ability to carry on with the daily occupation as well as sports.
The everyday lifestyle affects this physical fitness, in case you are not used used to "moving" in your daily life, your body will be sort of lazy and lack the power and ability to carry on with the daily occupation as well as sports.
In case your daily lifestyle involved motion, your body will be used to doing activities which will help improve your physical fitness.

Hope this helps :)

9 0

3 years ago

Read 2 more answers