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3 years ago

Why is it possible to remove a paper without moving the object on it.(apply newton's law)

Physics

1 answer:

klasskru [66]3 years ago

7 0

This is possible due to inertia of motion. which is nothing but newton's first law.

according to this law , an object tries to retain its state of motion or rest unless acted upon by an external force.

consider an object placed on a paper, initially both the object and paper are at rest. to pull the paper , we apply force on the paper and paper gains velocity. but the object keeps its motion of rest and hence the paper can be removed without moving the object.

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Please answer B, C, E, and D

Natasha_Volkova [10]

B- the biosphere needs the hydrosphere (water) to survive. They drink the water.
C- the biosphere use the géosphère as land/ home.
D-The géosphère depends on the hydrosphere for water. Without the water, the géosphère would eventually turn dry and nothing will be able to grow
E- The animals in the hydrosphere need air to survive. The gills on sea animals are used to filter out the water from the oxygen and use the oxygen to breathe. As the fish opens its mouth, water runs over the gills, and blood in the capillaries picks up oxygen that's dissolved in the water.

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1 year ago

Read 2 more answers

Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled wi

Korvikt [17]

Complete Question:

Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true?

Object T has a greater density than object B.
Object B has a greater density than object T.
Both objects have the same density.

Answer:

Object B has a greater density than object T

Explanation:

Any object partially or completely submerged in a liquid, experiments an upward force, equal to the weight of the volume displaced by the liquid. This force is called the buoyant force, and can be expressed as follows:

Fb = ρl * Vs*g

where ρl is the density of the liquid, and Vs is the submerged volume.

This force must be compared with the weight of the object, which is always downward, and can be expressed as follows:

Fg = ρb* Vb * g

where ρb, is the density of the object, and Vb is the total volume of the object, regardless which portion is submerged.

For object B, as it floats fully submerged, this means that both forces are equal in magnitude:

Fg = Fb⇒ ρb* Vb * g = ρl * Vs*g

As Vb = Vs (the object is fully submerged) this means that ρb =ρl.

For object T, as it floats partially submerged, this means that Fg < Fb:

Fg= ρt* Vt * g < Fb = ρl * Vs*g.

Now, we know that ρb =ρl, so we can replace in the equation above:

ρT* Vt * g < ρb*Vs*g

Simplifying common terms, and replacing Vs by KVt (where K is the fraction of the total volume which is submerged, i.e. K<1), we have:

ρt*Vt < ρb*K*Vt ⇒ ρt / ρb < K < 1 ⇒ ρt < ρb ⇒ ρb > ρt

3 0

2 years ago

51.Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveli

Komok [63]

Complete question is;

Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)

(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?

Answer:

A) P_out = 24 W

B) t = 1470 s

C) Q = 1140.72 KJ

Explanation:

We are given;

Input Power; P_in = 800 W

Efficiency; η = 3% = 0.03

A) Formula for efficiency is;

η = P_out/P_in

Making P_out the subject, we have;

P_out = η•P_in

P_out = 0.03 × 800

P_out = 24 W

B) We know that;

Power = work done/time taken

Thus;

P_out = mgh/t

We are given;

m = 3000 kg

h = 1.20 m

Thus, time is;

t = (3000 × 9.8 × 1.2)/24

t = 1470 s

C) amount of heat wasted is calculated from;

Q = (P_in - P_out)t

Q = (800 - 24) × 1470

Q = 1,140,720 J

Q = 1140.72 KJ

3 0

2 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

The material is peat, possibly.

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

2 years ago

How fast must a bug swim to keep up with the waves it produces? How fast must it move to produce a bow wave?

Elden [556K]

Answer:

A bug must swim as fast as the wave speed to keep up with the waves it produces. Moreso, a boat must be moving faster than the waves it creates to produce a bow wave.

6 0

3 years ago