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3 years ago

Draw a net force arrow on the picture below. What is the net force? State the direction.

Physics

2 answers:

zalisa [80]3 years ago

6 0

Answer:

Net force=0 N

Explanation:

The drawing will look like the attached image, with no net force on the object. We stablished a reference system located in the center of the body, with + axis take right and upward for x and y. All forces are on the x axis, there is one positive force of 7 N (pointing right) and to negative forces of -4 and -3 N (pointing left) if we add the vectors in the x axis we have that

+7-4-3=0 N

The net force is then 0 newton and the drawing has no net force vector then

exis [7]3 years ago

5 0

Answer: No arrow; no net force.

Explanation: -7 + 7= 0 same thing here

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A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago

A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

bixtya [17]

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

6 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

2 years ago

Which can be observed both on Earth and in an accelerating ship in space that is free from the effect of any gravitational field

Reil [10]

Answer:

apples falling from trees
people's feet touching the ground
sky divers moving toward the ground
balls bending downward after being thrown

Explanation:

When a space ship is accelerating in space, there is a force known as Inertia that kicks in. Inertia will mirror the effects of gravity on the ship even if there is no gravitational field effect such that anything that would happen where there is gravity, would continue to happen.

This means that apples will fall from trees, people's feet will touch the ground, sky divers will be pulled downwards and balls will bend downwards when thrown as well. These are the same effects expected on earth where gravity pulls things towards the earth's core.

8 0

2 years ago

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Which statement is TRUE?

Salsk061 [2.6K]

Statement B is true.

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3 years ago

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