Work is the ability to cause what?

A 30-g car rolls from a hill 12 cm high and is traveling at 154 cm/s as it travels along a 275 cm horizontal track. What is the

Bond [772]

Momentum of the car is 46.2 kgm/s

8 0

3 years ago

An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v

Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is $0.\overline {36}$ C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40 volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = $0.\overline {36}$ C

The charge on the second bulb, Q₂ = $0.\overline {36}$ C.

d. The question has left out parts

4 0

2 years ago

Which of the following statements is true of the strong nuclear force?

vredina [299]

Explanation:

it holds protons and neutrons together

5 0

3 years ago

power of a crane is 25000 watt calculate the time required by it to lift a load of 6000 kg up tp the height of 20 m

saul85 [17]

Solution:

We have,

Power [P] = 25000 Watt

Mass [m] = 6000 kg

Height [h] = 20 metres

Time [t] = ?

Now,

P = W/t = F x d/t = mxgx h/t

Or, 25000 = 6000 x 10 x 20/25000 [.......g = 10

m/s^2]

Or, t = 6000 x 10 x 20/25000

Or, t = 1200/25

Therefore, t = 48 second

Hence, the required time for the crane to lift the load is 48 seconds.

8 0

3 years ago

Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu

Elden [556K]

Answer:

Ex= -23.8 N/C Ey = 74.3 N/C

Explanation:

As the electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0 E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the cosine of θ, is just, by definition, the opposite of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y = 88.6 N/C + (-14.3 N/C) =74.3 N/C

7 0

3 years ago