2. Michelle bought 15.6 yards of fabric. She used 5.43 yards in her art 10 points

If you weighted 130 lbs on Earth how much would you weigh on the moon?

In-s [12.5K]

Answer:

21

Explanation:

Weight on the moon is 16.5 % of weight on earth

Weight on moon = 0.165 * 130

Weight on moon = 21 lbs

3 0

3 years ago

What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0

3 years ago

Scientific knowledge

emmasim [6.3K]

Answer:

D

Explanation:

As we progress and learn more, scientists make new discoveries that can contradict earlier discoveries, and since are technology is better today we are able to discover a lot more to add to or change our previous theories.

6 0

3 years ago

Read 2 more answers

What is R2 in the circuit?<br> WILL GIVE BRAINLIEST !!!!

Nataly_w [17]

Answer:

1. Rₑq = 4 Ω

2. R₂ = 6 Ω

3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V

4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A

Explanation:

1. Determination of the equivalent resistance

Voltage (V) = 12 V

Current (I) = 3 A

Resistance (Rₑq) =?

V= IRₑq

12 = 3 × Rₑq

Divide both side by 3

Rₑq = 12 / 3

Rₑq = 4 Ω

Thus, the equivalent resistance (Rₑq) = 4 Ω

2. Determination of R₂.

Equivalent resistance (Rₑq) = 4 Ω

Resistance 1 (R₁) = 12 Ω

Resistance 2 (R₂)

Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:

Rₑq = R₁ × R₂ / R₁ + R₂

4 = 12 × R₂ / 12 + R₂

Cross multiply

4(12 + R₂) = 12R₂

48 + 4R₂ = 12R₂

Collect like terms

48 = 12R₂ – 4R₂

48 = 8R₂

Divide both side by 8

R₂ = 48 / 8

R₂ = 6 Ω

3. Determination of the total voltage (Vₜ), V₁ and V₂.

From the question given above, the total voltage is 12 V

Since the resistors are arranged in parallel connection, the same voltage will go through them.

Thus,

Vₜ = V₁ = V₂ = 12 V

4. Determination of the total current (Iₜ), I₁ and I₂

From the question given above, the total current (Iₜ) is 3 A

Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.

Vₜ = V₁ = 12 V

R₁ = 12 Ω

I₁ =?

V₁ = I₁R₁

12 = I₁ ×12

Divide both side by 12

I₁ = 12 / 12

I₁ = 1 A

Next, we shall determine I₂. This can be obtained as follow:

Iₜ = 3 A

I₁ = 1 A

I₂ =?

Iₜ = I₁ + I₂

3 = 1 + I₂

Collect like terms

I₂ = 3 – 1

I₂ = 2 A

5 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago