All categories

3 years ago

Newton’s second law of motion states that when a force acts on an object, the object’s acceleration is equal to _____.

1 answer:

3 0

Newton’s second law of motion states that when a force acts on an object, the object’s acceleration is equal to the net force divided by the object’s mass, and is in the same direction as the net force.

You might be interested in

A force of 35 n acts on an object which has a mass of 5.4 kg. what acceleration (in m/s2) is produced by the force

galina1969 [7]

∑F = ma
a = ∑F/m
a = 35 N / 5.4 kg
a = 6.5 m/s²

4 0

3 years ago

A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

And now we can re-arrange the eq.(1) to solve for the resistivity:

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

5 0

3 years ago

What do parentheses mean in a Calc formula? A. Ignore the numbers in parentheses until the end. B. Calculate the numbers in pare

kirill115 [55]

The answer is B.........

3 0

3 years ago

Read 2 more answers

4. Premature wrinkling due to overexposure to the Sun, what am I?

Irina-Kira [14]

Premature Aging

Mark as brainlist

5 0

2 years ago

Why do liquids solidefy?

Setler79 [48]

energy extracted out of liquids an atoms are left to come closer arrange themselves shorter distance and then they solidify

4 0

3 years ago