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3 years ago

A train has a constant velocity of 2 m/s. east what is the magnitude of the horizontal acceleration of the trian?

Physics

1 answer:

kupik [55]3 years ago

4 0

Answer:

0 m/s²

Explanation:

The velocity is constant, so there is no acceleration.

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a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin

creativ13 [48]

If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

(2.5 x 1 newton) = 2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

(2.5 x 120 newtons) = 300 newtons.

4 0

3 years ago

If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?

fiasKO [112]

1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J

4 0

2 years ago

An engineering firm is designing a ski lift. The wire rope needs to travel with a linear velocity of 2.0 meters per second, and

Mazyrski [523]

Answer:

The diameter of the bull-wheel is 3.82

Explanation:

Given that,

Velocity = 2.0 m/s

Angular velocity = 10 rev/m

$\omega=10\times\dfrac{2\pi}{60}$

$\omega=1.0472\ rad/s$

We need to calculate the diameter of bull-wheel

Using formula of angular velocity

$v= r\omega$

$r=\dfrac{v}{\omega}$

Put the value into the formula

$r=\dfrac{2.0}{1.0472}$

$r=1.91\ m$

The diameter of the bull-wheel

$D=2r$

$D=2\times1.91$

$D=3.82\ m$

Hence, The diameter of the bull-wheel is 3.82 m.

6 0

3 years ago

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

natka813 [3]

If 56.5kJ are needed to raise the temp by 90°C and if the heater is 60% efficient that means that:
60% X y = 56.5kJ
where y is the electrical energy in kJ that the heater will use.
y = 94.2kJ

6 0

2 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

8 0

3 years ago