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3 years ago

Newton’s laws do not apply to small objects?

Physics

1 answer:

Soloha48 [4]3 years ago

7 0

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

8 0

3 years ago

A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part

borishaifa [10]

Answer:

v₂ = 7/ (0.5)= 14 m/s

Explanation:

Flow rate of the fluid

Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.

The formula for calculated the flow rate is:

Q= v*A Formula (1)

Where :

Q is the Flow rate (m³/s)

A is the cross sectional area of a section of the pipe (m²)

v is the speed of the fluid in that section (m/s)

Equation of continuity

The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:

Q₁= Q₂

Data

A₁ = 2m² : cross sectional area 1

v₁ = 3.5 m/s : fluid speed through A₁

A₂ = 0.5 m² : cross sectional area 2

Calculation of the fluid speed through A₂

We aply the equation of continuity:

Q₁= Q₂

We aply the equation of Formula (1):

v₁*A₁= v₂*A₂

We replace data

(3.5)*(2)= v₂*(0.5)

7 = v₂*(0.5)

v₂ = 7/ (0.5)

v₂ = 14 m/s

4 0

3 years ago

What is the farthest distance parallaxes can be used to measure star distances from Earth?

JulsSmile [24]

Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.

8 0

3 years ago

How far can your little brother get if he can travel at 2.5 m/s and in 5?

KatRina [158]

d=? v=2.5 u=0 and t=5 therefore the formula to be used to find the distance my brother covered is d=1/2(v-u)t

d=1/2(2.5-0)5

=6.15m

4 0

2 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0

3 years ago