The weight of an object is the force pulling the object:

What needs to be done to this Circuit so that the lamp lights up

UkoKoshka [18]

Explanation:

the correct answer is Option C, add a cell and close the switch.

hope this helps you.

5 0

3 years ago

In deep space, sphere A of mass 94 kg is located at the origin of an x axis and sphere B of mass 100 kg is located on the axis a

vlabodo [156]

(a) $-3.48\cdot 10^{-7} J$

The gravitational potential energy of the two-sphere system is given by

$U=-\frac{Gm_A m_B}{r}$ (1)

where

G is the gravitational constant

$m_A = 94 kg$ is the mass of sphere A

$m_B = 100 kg$ is the mass of sphere B

r = 1.8 m is the distance between the two spheres

Substitutign data in the formula, we find

$U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.8 m}=-3.48\cdot 10^{-7} J$

and the sign is negative since gravity is an attractive force.

(b) $1.74\cdot 10^{-7}J$

According to the law of conservation of energy, the kinetic energy gained by sphere B will be equal to the change in gravitational potential energy of the system:

$K_f = U_i - U_f$ (2)

where

$U_i=-3.48\cdot 10^{-7} J$ is the initial potential energy

The final potential energy can be found by substituting

r = 1.80 m -0.60 m=1.20 m

inside the equation (1):

U=-\frac{(6.67\cdot 10^{-11})(94 kg)(100 kg)}{1.2 m}=-5.22\cdot 10^{-7} J

So now we can use eq.(2) to find the kinetic energy of sphere B:

$K_f = -3.48\cdot 10^{-7}J-(-5.22\cdot 10^{-7} J)=1.74\cdot 10^{-7}J$

4 0

3 years ago

A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1

Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

8 0

3 years ago

In two or more complete sentences, explain how you can prove that the number of degrees that the Moon rotates around the Earth e

PolarNik [594]

Answer:

Follows are the explanation to this question:

Explanation:

In this solution, it is defined that there are two principal motions for the moon, which are its revolution as well as rotation. In such a movement called revolution, its Moon is relocating around the Earth, in which the approximate movement of the moon from around earth has an average movement of about 13.2° per day, or 92 degrees every week, that's once in 27.3 days.

3 0

3 years ago

Read 2 more answers

Someone help me with letter d. and e. I’m supposed to solve for y.

Bumek [7]

Answer:

$t=\frac{a}{\sqrt{1-\frac{y^2}{c^2}}}\\\frac{y^2}{c^2}=1-\frac{a^2}{t^2}\\y^2=\frac{c^2t^2-c^2a^2}{t^2}\\y=\frac{c}{t}\sqrt{t^2-a^2}$

$t=\frac{a}{\sqrt{1-\frac{v^2}{y^2}}}\\\frac{v^2}{y^2}=1-\frac{a^2}{t^2}\\\\\frac{v^2}{y^2}=\frac{t^2-a^2}{t^2}\\y^2=\frac{v^2t^2}{t^2-a^2}\\y=\frac{vt}{\sqrt{t^2-a^2}}$

7 0

3 years ago