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3 years ago

Which would fall with a greater acceleration in a vacuum-a leaf or a stone?

Physics

2 answers:

Anarel [89]3 years ago

6 0

Since a vacuum has no air resistance they would both fall at the same rate

hichkok12 [17]3 years ago

4 0

They would fall with the same acceleration

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a person was using a spanner to loosen a tight nut.Their mass was 50kg the spanner was 0.5m long i) what is the persons weight?

sineoko [7]

1) Weight of the person: 490 N

2) Maximum torque: 245 Nm

Explanation:

The weight of a body is equal to the gravitational force exerted on the body; it is given by the equation

$F=mg$

where

m is the mass of the body

g is the acceleration due to gravity

For the person in this problem,

m = 50 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity on Earth's surface

Therefore, the weight of the person is

$F=(50)(9.8)=490 N$

The turning force (also called torque) exerted by a force rotating an object is given by

$\tau = Fd sin \theta$

where

F is the magnitude of the force

d is the length of the arm (the distance between the force and the pivotal point)

$\theta$ is the angle between the direction of the force and the arm

For the spanner in this problem,

F = 490 N is the force applied (the weight of the person)

d = 0.5 m is the arm (the length of the spanner)

The maximum torque is obtained when $\theta=90^{\circ}$ , therefore it is:

$\tau=(490)(0.5)(sin 90^{\circ})=245 N\cdot m$

Learn more about weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

About torque:

brainly.com/question/5352966

#LearnwithBrainly

5 0

3 years ago

Which of the following is NOT true about a pool of water and a piece of ice? A They have the same composition. B They are in dif

anastassius [24]

C. They have different chemical properties.

7 0

3 years ago

Read 2 more answers

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$