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3 years ago

What is the acceleration of an object in free fall near Earth's surface?

Physics

2 answers:

Vitek1552 [10]3 years ago

8 0

Answer:

D 9.8 m/s^2

Explanation:

The force of gravitational gravity on earth is 9.8 m/s^2

Licemer1 [7]3 years ago

8 0

D) 9.8 m/(s^2) is the approximate acceleration of a free falling object near Earth’s surface

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Help me I do not understand

ololo11 [35]

Answer:

170N

Explanation:

First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N

4 0

3 years ago

I wish to use a step up transformer to turn an initial RMS AC voltage of 100 V into a final RMS AC voltage of 200 V. What is the

zhuklara [117]

Answer:

1:2

Explanation:

It is given that,

Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.

We need to find the ratio of the number of turns in the primary to the secondary for step up transformer.

For a transformer, $\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}$

So,

$\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}$

So, the ratio of the number of turns in the primary to the secondary is 1:2.

4 0

3 years ago

A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro

ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let $u_1, u_2 , v_f$ be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

$m=2.50\times 10^{4}$

Conserving momentum

$mu_1+3mu_2=4mv_f$

$4m+6m=4mv_f$

$v_f=2.5 m/s$

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

$Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J$

$Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J$

$Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ$

4 0

3 years ago

Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago

The mixture you separated was mixture oof iron filings, sand, and salt. Based on your understanding of matter, is this mixture a

scZoUnD [109]

<span>The answer is B, because a heterogeneous mixture consists of elements that have the same appearance and composition, If sand, salt, and iron filings are combined, they would look like one uniform element rather than three separate elements.</span>

3 0

3 years ago

Read 2 more answers