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2 years ago

If a swimmer pushes off a pool wall with a force of 250n, what is her acceleration.

Physics

1 answer:

grigory [225]2 years ago

3 0

Her acceleration is

(250) divided by (the swimmer's mass, in kilograms).

The unit is " meters per second² " .

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The distance between the centers of the wheels of a motorcycle is 146 cm. The center of mass ofthe motorcycle, including the rid

Vitek1552 [10]

Answer:

9.12267515924 m/s²

Explanation:

Here the moment created by the wheels and the moment created by the center of gravity will balance each other.

h = Height of the center of mass = 78.5 cm

d = Distance from back wheel to the center of mass = $\dfrac{146\times 10^{-2}}{2}\ m$

g = Acceleration due to gravity = 9.81 m/s²

a = Horizontal acceleration

The equation is of the form

$mgd=Fh\\\Rightarrow mgd=mah\\\Rightarrow a=\dfrac{gd}{h}\\\Rightarrow a=\dfrac{9.81\times \dfrac{146\times 10^{-2}}{2}}{78.5\times 10^{-2}}\\\Rightarrow a=9.12267515924\ m/s^2$

The horizontal acceleration of the motorcycle that will make the front wheel rise off the ground is 9.12267515924 m/s²

8 0

3 years ago

What is the weight of a box with a mass of 150 kg on Earth?

Dmitry [639]

Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

6 0

1 year 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

2 years ago

To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length

san4es73 [151]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

3 0

2 years ago

Which question asks for an opinion?

blsea [12.9K]

Answer:

Explanation:

It says is it a good idea the person so 1 person can say no and the other one can say yes so it is asks for a opinion

5 0

3 years ago