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2 years ago

What two factors affect the rate of acceleration in an object

Physics

2 answers:

Leto [7]2 years ago

8 0

Newton's second law states that F=ma so that means force =mass×acceleration force and mass

jonny [76]2 years ago

4 0

Force and mass affect acceleration
F=ma

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What do we mean by the observable universe?

mote1985 [20]

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2 years ago

Activity 1 more or less, pls answer this ASAP

AURORKA [14]

Answer:

ididate is a good one and

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2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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2 years ago

An electron is emmited by an atomic nucleus in the process of...

Sliva [168]

I’m pretty sure it’s Radioactive decay

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2 years ago

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What is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C ?

torisob [31]

0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.

<h3 /><h3>What is a resistor?</h3>

A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.

The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:

R = R0[1+ alpha(T-T0)] where:

R is the new resistance we are looking for

alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x $10^{-4}$ (1/°c)

T0 is the standard temperature =25.8°C

R0 is the resistance at T0 = 0.0200 ohms

T is the temperature at which we want to get R = 0

Substitute in the equation to get R as follows:

R = 0.0200 [1+( ₋ 4.8 x $10^{-4}$ ) (0-25.8)] = 0.02020 ohm

To know more about resistance refer to: brainly.com/question/11431009

#SPJ1

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11 months ago