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2 years ago

Calculate the time it would take a motor with a power of 0.6kW to lift a suitcase of 20kg a distance of 50cm

Physics

1 answer:

True [87]2 years ago

7 0

Answer:

0.163 s

Explanation:

Appying,

P = mgh/t................ Equation 1

Where P = power of the motor, m = mass of the suitcase, h = vertical distance, t = time, g = acceleration due to gravity.

make t the subject of the equation,

t = mgh/P................ Equation 2

Given: m = 20 kg, h = 50 cm = 0.5 m, P = 0.6 kW = 600 W

Constant: g = 9.8 m/s²

Substitute these values into equation 2

t = (20×0.5×9.8)/600

t = 0.163 s

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velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

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3 years ago

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Answer:

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Explanation:

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3 years ago

The yellow star has much less mass the blue star and so who will live longer

In-s [12.5K]

The yellow star will live longer as it has less mass

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3 years ago

Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1

andriy [413]

Answer:

$\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.$

Explanation:

Given:

Mass, $\rm m\pm\Delta m = 1.3\pm 0.4\ kg.$

Velocity, $\rm v\pm \Delta v = 5.2\pm 0.2\ m/s.$

where,

$\rm \Delta m,\ \Delta v$ are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

$\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.$

The uncertainty in kinetic energy is given as:

$\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.$

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3 years ago

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Answer:

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Explanation:

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3 years ago