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3 years ago

An object with little mass won't require a lot of force to move.

Physics

1 answer:

Anna35 [415]3 years ago

7 0

Answer:

True

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.

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What is kinetic energy? Question 1 options: energy that is due to motion energy that is stored the mass of an object the rate at

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Energy that is due to motion

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3 years ago

During a certain time interval, a constant force delivers an average power of 4 watts to an object. if the object has an average

bija089 [108]

<span>Force = Work done / distance = 4Nm / 2m = 2N</span>

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3 years ago

Read 2 more answers

Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to

Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength= $\frac{velocity of sound }{frequency}$

= $\frac{343}{78}$ =4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

$x-\left ( 8.2-x\right )=4.397$ for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

$x-\left ( 8.2-x\right )=0$

2x=8.2

x=4.1m

Third

$\left ( 8.2-x\right )-x=4.397$

x=1.9015 m

6 0

3 years ago

An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m

dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, $F=1.85\times 10^{-15}\ N$

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

$F=B\times q\times v$

q = charge on an electron, $q=1.6\times 10^{-19}\ C$

v = velocity of an electron

$v=\dfrac{F}{Bq}$

$v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}$

v = 34007.35 m/s

Hence, this is the required solution.

4 0

3 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago