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2 years ago

Dante uses 14 J of work to lift a weight for 30 seconds. How much power did he use?

Physics

1 answer:

kirill [66]2 years ago

3 0

Answer:

The answer is 0.47 watts

power is calculated by Work/Time

14/30 =0.46666666 which is approximately 0.47watts

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(YOU WILL GET BRAINLIEST)Matter may be classified as a pure substance or a mixture. Where on the Venn diagram would you insert t

inessss [21]

I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms

3 0

1 year ago

Read 2 more answers

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

wolverine [178]

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

$\omega =$ Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

$1000J > \frac{1}{2}mv^2$

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2$

$1000J = \frac{1}{2}mv^2$

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

6 0

1 year ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

1 year ago

Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman

Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2$

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0

1 year ago

Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes

Arte-miy333 [17]

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, $v_1=65\ km/h$

Radius of the circular arc, $r_1=95\ m$

Car 2 has twice the speed of Car 1, $v_2=130\ km/h$

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

According to given condition,

$\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}$

$\dfrac{65^2}{95}=\dfrac{130^2}{r_2}$

On solving we get :

$r_2=380\ m$

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

4 0

2 years ago