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3 years ago

14

A 200-kilogram wood crate is pushed across a wood floor How much force will be needed to move the box at a constant velocity of

0.5 m/s

1 answer:

lakkis [162]3 years ago

4 0

Answer:

100N

Explanation:

F = m x a

m = 200kg

a = 0.5m/s

F = 200 x 0.5

F = 100

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A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th

DiKsa [7]

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

= 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= $\frac{0.4937}{55\times10^{-6}}$

Density of block = 8977.7 kg/m^3

4 0

3 years ago

A person is riding on a Ferris wheel. When the wheel makes one complete turn, the net work done on the person by the gravitation

lara31 [8.8K]

Answer:

0

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

d = Vertical height from the ground

F = Force = Weight = mg

Net work done would be

$W_n=W_{up}+W_{down}\\\Rightarrow W_n=Fdcos180+Fdcos0\\\Rightarrow W_n=-mgd+mgd\\\Rightarrow W_n=0$

Hence, the work done on the person by the gravitational force is 0

7 0

3 years ago

rjkz [21]

C.<span>a stable internal attribution</span>

4 0

3 years ago

Read 2 more answers

You place a solid cylinder of mass M on a ramp that is inclined at an angle β to the horizontal. The coefficient of static frict

zlopas [31]

Answer:

Explanation:

Let the frictional force required be f .

frictional force is responsible in creating rotational motion in the cylinder.

torque created by frictional force = f R

if angular acceleration be α

α = f R / I , I is moment of inertia of cylinder .

α = a / R , a is linear acceleration.

f R / I = a / R

a = f R² / I

linear acceleration a of cylinder down the slope

ma = mgsinθ - f , ( f force is acting upwards and mgsinθ is acting downwards )

mf R² / I = mgsinθ -f

f ( m R² / I + 1) = mgsinθ

f = mgsinθ / ( m R² / I + 1)

= mgsinθ / ( m R² / mk² + 1) , k is radius of gyration of cylinder.

= mgsinθ / ( R² / k² + 1)

Putting the given values

f = Mgsinβ /( R² / k² + 1)

for cylinder , R² / k² = 2

f = Mgsinβ /3

6 0

3 years ago

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

julia-pushkina [17]

Answer:

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

$V=\pi r^2l$

$V=\pi \times 0.5^2\times 2 \ \mu m^3$

$V=1.57 \ \mu m^3$

Surface area,A

A=π d l

$A=\pi \times 1 \times 2\ \mu m^2$

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

4 0

3 years ago