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3 years ago

What is amperage?

Physics

1 answer:

Shtirlitz [24]3 years ago

3 0

question one b

question 2 i think a

3 d

4 c

5 not sure but wanting to say d

6 letter b

7 not sure

8 idk

9 i have no idea

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For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m

Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is $6.85\times 10^{5}$ newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

$\Sigma F = F-R = 0$ (Eq. 1)

Where:

$F$ - Forward trust, measured in newtons.

$R$ - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: ( $F = 6.85\times 10^{5}\,N$ )

$R = F$

$R = 6.85\times 10^{5}\,N$

The magnitude of the resistive force exerted by the water is $6.85\times 10^{5}$ newtons.

4 0

3 years ago

What is the SI unit for momentum?

serg [7]

It’s a vector quantity, which means it possesses both magnitude and direction. So the SI unit would be B)kg•m/s

4 0

3 years ago

Read 2 more answers

1. A brick and a peanut are dropped of the roof of a house at the same time, which

iris [78.8K]

Answer:

It would be the brick since the mass and the weight is greater than the peanut I might be wrong on this but thats the best answer i can give

Explanation:

3 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

2. The light from the Sun reaches Earth in 8.3 minutes. The speed of light is 11,184,681 miles/minute. How far is

12345 [234]

Answer:

92,832,852.3 miles

Explanation:

6 0

2 years ago