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Sedbober [7]
3 years ago
13

A robot is exploring​ charon, the dwarf planet​ pluto's largest moon. Gravity on charon is 0.278 meters per second ​[m divided b

y s squared​]. During its​ investigations, the robot picks up a small spherical rock for inspection. The rock has a diameter of 6 centimeters​ [cm] and is lifted 15 centimeters​ [cm] above the surface. The specific gravity of the rock is 10.8. The mechanism lifting the rock is powered by a 10​-volt ​[v] power supply and draws 1.83 milliamperes​ [ma] of current. It requires 12 seconds​ [s] to perform this lifting task. What is the efficiency of the​ robot
Physics
1 answer:
DochEvi [55]3 years ago
5 0

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

V = \frac{4}{3}\pi r^3

V = \frac{4}{3}\pi *(\frac{0.06}{2})^3

V = 1.13 * 10^{-4} m^3

Now it is given that it's specific gravity is 10.8

So density of rock is

\rho = 10.8 * 10^3 kg/m^3

mass of the stone will be

m = \rho V

m = 10.8* 10^3 * 1.13 * 10^{-4}

m = 1.22 kg

now change in potential energy is given as

\Delta U = mgH

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

\Delta U = 1.22* 0.278 * 0.15

\Delta U = 0.051 J

Now energy supplied by internal circuit of robot is given by

E = Vit

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

E = 10* 1.83 * 10^{-3} * 12

E = 0.22 J

Now efficiency is defined as the ratio of output work with given amount of energy used

\eta = \frac{\Delta U}{E}*100

\eta = \frac{0.051}{0.22} = 0.23

so efficiency will be 23 %

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A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0
3 years ago
A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict
Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

F=157\times 0.19\\\\F=29.83\ N

(B) The force is simply given by :

F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

6 0
3 years ago
A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio
Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0   The car eventually does stop.

vi = 72 km/hr * [ 1000 m/  km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 =   a

a = - 5 m/s^2   The minus sign tells you the vehicle is slowing down. It sure should be.

Force = m * a

F = - 850 * (-5)

F =  - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0
3 years ago
you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t
telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

  • velocity of the ant, v = 2 m/s
  • change in position of the ant, Δx = 0.81 m
  • initial time, t₁ = 4 s
  • time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s

The time elapsed since you stopped the stopwatch is calculated as;

t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

5 0
3 years ago
Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
3 years ago
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