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3 years ago

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Physics

1 answer:

Triss [41]3 years ago

4 0

The acceleration of the boat is $2 m/s^2$ (eastward)

Explanation:

We can solve this problem by using Newton's second law, which states that:

F = ma (1)

where

F is the net force on a body

m is its mass

a is its acceleration

First of all, we have to find the net force acting on the boat. We have:

- The force of the motor, 100 N eastward

- The force of the air resistance, 60 N westward

So the net force is:

F = 100 N - 60 N = 40 N (eastward)

Now we can apply eq.(1), using:

F = 40 N

m = 20 kg (mass of the boat)

and solving for a, we find the acceleration:

$a=\frac{F}{m}=\frac{40}{20}=2.0 m/s^2$

And the direction is the same as the net force (eastward).

Learn more about acceleration and Newton laws of motion:

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The radius of a small ball is around 3.79747 cm. The radius of a basketball is about 3.16 times larger. What is the ratio of the

Svetradugi [14.3K]

Explanation:

The ratio of the areas is the square of the ratio of the radii.

A/A = 3.16² = 9.99

The ratio of the volumes is the cube of the ratio of the radii.

V/V = 3.16³ = 31.6

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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

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

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What is the momentum of a 200 kg truck travelling at 20 m/s?

geniusboy [140]

Answer:

p = 4000 kg-m/s

Explanation:

Given that,

The mass of a truck, m = 200 kg

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We need to find the momentum of the truck. The formula for momentum is given by :

p = mv

so,

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Answer:

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Explanation:

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