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4 years ago

Holding a book while walking means i am doing work on the book. true or false

Physics

1 answer:

Ulleksa [173]4 years ago

6 0

False, while holding the book it means there is no work done on the book.

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PLEASE HELP!! PHYSICS QUESTION-DOES ANYONE KNOW HOW TO DO THIS WITH THE WORKING OUT?!?

notsponge [240]

Answer:

$1.5 \Omega$

Explanation:

The resistance of a piece of wire is given by

$R=\frac{\rho L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area

In this problem, we have

$\rho = 3.3\cdot 10^{7} m$

L = 7.0 m

The diameter of the wire is 0.14 cm, so the radius is 0.07 cm, therefore the cross sectional area is

$A=\pi r^2 = \pi (0.07)^2=0.0154 cm^2 = 0.0154\cdot 10^{-4} m^2$

Therefore, the resistance is

$R=\frac{(3.3\cdot 10^{-7})(7.0)}{(0.0154\cdot 10^{-4})}=1.5 \Omega$

4 0

3 years ago

A 2.00 kg cart on a frictionless track is pulled by force of 3.00 N. What is the acceleration of the cart?

kow [346]

1.5 will be its acceleration

5 0

3 years ago

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

7 0

3 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

4 0

4 years ago

Is this the actual answer??

Nastasia [14]

It probably is the actual answer.

3 0

3 years ago