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1 year ago

How can u create a fair test? Short answer only:

Physics

2 answers:

omeli [17]1 year ago

7 0

Answer:

A way you can create a fair test by making sure that you change one factor at a time while keeping all other conditions the same.

Explanation:

Hope it helps! Have a great day!

Lilac~

Crazy boy [7]1 year ago

5 0

Answer:

making sure that you change one factor at a time while keeping all other conditions the same

Explanation:

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A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart

Ber [7]

Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
$\sum F = ma$
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force $F_f$ acting in the opposite direction. So Newton's second law can be rewritten as
$F-F_a = ma$
or
$F=ma + F_f$

since the frictional force is 15 N and we want to achieve an acceleration of $a=1.50 m/s^2$ , we can substitute these values to find what is the force the man needs:
$F=(30 kg)(1.5 m/s^2)+15 N=60 N$

8 0

3 years ago

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

0.97c

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = 0.97c

3 0

2 years ago

What is magnetism please define it

Scrat [10]

Hello There!

It's a Property of matter where atoms in an object are aligned into domains

4 0

3 years ago

A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.

MaRussiya [10]

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm= $\frac{20}{100}=0.20m$

1 m=100 cm

Mass of rod,M=190 g= $\frac{190}{1000}=0.19kg$

Mass of ball,m=19 g= $\frac{19}{1000}=0.019 kg$

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

$I_{rod-clay}=I_{rod}+I_{clay}$

$I_{rod-clay}=\frac{1}{3}ML^2+mL^2$

Substitute the values then we get

$I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2$

$I_{rod-clay}=3.29\times 10^{-3} kgm^2$

Now, the center of mass of the combination of the rod and clay is given by

$y=\frac{Md_1+md_2}{M+m}$

Substitute $d_1=\frac{L}{2}$ =Distance between pivot and the center of the rod

$d_2=L=$ The distance between rod and clay

Using the formula

$y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}$

$y=0.1091$ m

Time period of the oscillation of the system of the rod and the clay is given by

$T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}$

g= $9.8m/s^2$

Using the formula

Time period= $2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}$

Time-period=0.76 s

Hence, the period =0.76 s

8 0

3 years ago

A coin is placed 10.8 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly incr

Advocard [28]

Answer:

Coefficient of static friction = 0.547

Explanation:

The maximum resistive force offered by the body(coin)against the applied force(turntable) to continue its state of motion is called Coefficient Of Static Friction

When given

The angular velocity which is 51.1rpm

We are going to calculate velocity of coin

= r*w(angular velocity) = 0.108*(51.1*2π/60) = 0.58

Coefficient=F/N

F= coefficient* N = coefficient*m*g

F= mv^2/r which is centripetal force

Therefore, mv^2/r = coefficient*m*g

coefficient = mv^2/mgr = v^2/rg

Where g,gravity = 9.81

Substitute into the equation coefficient = 0.58^2/9.81*0.108

= 0.547

7 0

3 years ago