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4 years ago

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Answer will give you brainliest and 50 pts An object with a mass of 2000 g accelerates 8.3 m/s² when an unknown force is applied

to it. What is the amount of the force? Show work

1 answer:

Natali [406]4 years ago

4 0

Force = (mass) x (acceleration)

Mass = 2000 grams = 2 kilograms

Force = (2 kg) x (8.3 m/s²)

Force = <em>16.6 Newtons</em>

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A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

lions [1.4K]

Answer:

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

(c) 1235.06N and 252.05 as a ratio

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity

Making $u_{i}$ the subject we obtain $u_{i}=\frac {u_{y}}{sin \theta}$

Since $u_{y}$ and $\theta$ are known as $46.43ms^{-1}$ and $80^{o}$ respectively, then $u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}$

Therefore, the velocity of potato is $47.15ms^{-1}$

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation $v^{2}=u^{2}+2sa$

The final velocity v is answer in part a which is $47.15ms^{-1}$ , initial velocity u is $0ms^{-1}$ hence the equation is re-written as

$v^{2}=2sa$ and making a the subject we obtain

$a=\frac {v^{2}}{2s}$

$a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}$

Therefore, average acceleration is $2470.13ms^{-2}$

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is $2470.13ms^{-2}$

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

$R=\frac {F}{W}=\frac {F}{mg}$ and since F=1235.06, m=0.500kg and g=9.8

$R=\frac {1235.06}{0.500*9.8}$ =252.05

Therefore, ratio of average force to weight is 252.05

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A toy helicopter takes off and moves 6 m up and then 1 m back down. What

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the answer is 5 m up AKA "B"

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Why are experiments often performed in laboratories?

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3 years ago

Read 2 more answers

A horizontal force of 25 N is required to push a wagon across a sidewalk at a constant speed.

lions [1.4K]

Answer:

a. The net force acting on the wagon is zero

b. The value of the force of friction is 25 N

c. The effect would be is the frictional force must be increase

Explanation:

Lets explain how to solve the problem

A horizontal force of 25 N is required to push a wagon across a sidewalk

at a constant speed

a.

→ Constant speed means acceleration is zero

→ Force = mass × acceleration

→ Force = mass × 0

→ Force = 0

<em>The net force acting on the wagon is zero</em>

b.

The wagon moves with constant speed, then

→ ∑ force = 0

There are two forces acting on the wagon external force of 25 N,

and the frictional force which is opposite to the direction of motion

→ 25 - Frictional force = 0

→ Frictional force = 25 N

<em>The value of the force of friction is 25 N</em>

c.

The force increased to 30 N

If the wagon still moves with constant speed

Due to Newton's law ∑ force in the direction of motion = mass ×

acceleration, but acceleration is 0 because the wagon moves with

constant speed, then ∑ F = 0

So the frictional force must increase to 30 N

<em>The effect would be is the frictional force must be increase</em>

<em></em>

<em>If the frictional force does not increase the wagon will accelerate</em>

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4 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

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