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3 years ago

What is the power involved in lifting a 10-kg object 1.0 m in 0.50 s?

1 answer:

5 0

In this question, we know that mass= 10 kg = 10 x 1000 = 10,000 g
Distance = 1 m and Time = 0.5 s

Power = Force x Velocity
Velocity = Distance / Time = 1 m / 0.5 s = 2 m/s
So, Power = Force x (Distance / Time)
But Force= Mass x Acceleration due to gravity (g)
So, Force = 10 kg x 9.8 m/s $^{2}$ = 98
Therefore, Power =Force x Velocity= 98 x 2 = 196 W

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M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.

MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V is $6.49\times10^6m/s$

<h3>How to calculate the speed of the electron?</h3>

We know, that the energy of the system is always conserved.

Using the Law of Conservation of energy,

$\triangle K+\triangle$ U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

$(\frac{1}{2} mv^2)+(-q\triangle V)$ =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let $v_f$ and $v_i$ represent the final and initial speed.

Here, $v_i$ =0

Solving for $v_f$ , we get:

$v_f=\sqrt{\frac{-2q\triangle V}{m} }$

$=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }$

=6.49 $\times10^6$ m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

4 0

1 year ago

Worth 15 points<br><br> Please answer which one matches for each conversion

FrozenT [24]

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

5 0

2 years ago

If T, = 40 N, find T, and the mass of the weight (W).

seraphim [82]

Answer:4kg

Explanation:

acceleration due to gravity(g)=10m/s^2

Weight(w)=40N

Weight=mass x g

40=mass x 10

Divide both sides by 10

Mass =40/10

Mass=4kg

5 0

3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

How much force is needed to accelerate a 1,500 kg car at a rate of 8 m/s2?

lesya [120]

Force = mass * acceleration = 1500kg * 8m/s²

3 0

3 years ago