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SVETLANKA909090 [29]

3 years ago

6

What did Benjamin Thompson contribute to science? A. P(↓1)V(↓1)=P(↓2)V(↓2) B. V(↓1)/T(↓1)=V(↓2)/T(↓2) C. the idea of heat energy

D. the Kinetic Theory of Matter?

1 answer:

spin [16.1K]3 years ago

8 0

C.is the answer the idea of heat energy.

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A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

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A reference point would be something not on the ship which could be used to calculate distance traveled.

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3 years ago

Read 2 more answers

An electron and a proton are each placed at rest in an electric field of 500 N/C. Calculate the speed (and indicate the directio

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Answer:

For proton: 2592 m/s In the same direction of electric field.

For electron: 4752000 m/s In the opposite direction of electric field.

Explanation:

E = 500 N/C, t = 54 ns = 54 x 10^-9 s,

Acceleration = Force /mass

Acceleration of proton, ap = q E / mp

ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2

Acceleration of electron, ae = q E / me

ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2

For proton:

u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.

For electron:

u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.

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Answer:

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A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

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(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

$\tau = F r$

where

F is the magnitude of the thrust

r is the radius of the wire

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F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

$\tau = (0.795 N)(30.9 m)=24.6 N m$

(b) $0.035 rad/s^2$

The equivalent of Newton's second law for a rotational motion is

$\tau = I \alpha$

where

$\tau$ is the torque

I is the moment of inertia

$\alpha$ is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

$I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2$

And so we can solve the previous equation to find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2$

(c) $1.08 m/s^2$

The linear acceleration (tangential acceleration) in a rotational motion is given by

$a=\alpha r$

where in this problem we have

$\alpha = 0.035 rad/s^2$ is the angular acceleration

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Substituting the values, we find

$a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2$

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Answer: B

Explanation:

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