Ask question

Ask question

All categories

3 years ago

5

When is the ideal time to take a resting heart rate? O A. After exercise OB. After a meal OC. Hirst thing in the morning D. Befo

re going to bed

1 answer:

stira [4]3 years ago

7 0

Answer:

D

Explanation:

You might be interested in

A lamp draws a current of 0.1 A when it is connected to a 122-V source. (

Setler79 [48]

The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W

4 0

3 years ago

In an economy, the demand for labor is given by the equation W = 15 - (1/200) L and the supply of labor is given by the equation

mr_godi [17]

Answer:

the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

Explanation:

The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

$W_{Demand} = W_{Supply}$

15 - (1/200) L = 5 + (1/200) L

15 - 5 = (1/200) L + (1/200) L

10 = (2/200) L

(10*200)/2 = L

1000 = L

Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:

W = 15 - (1/200) L

W = 15 - (1/200) 1000

W = 10

Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

7 0

3 years ago

The measure of the force with which air molecules push on a surface is called

alex41 [277]

It is called air pressure

7 0

3 years ago

Two cars, one of mass 1400 kg, and the

Vikki [24]

Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector

v₁ = (14 m/s) i

and the second car is moving upward so that its velocity vector is

v₂ = (20 m/s) j

Then the total momentum of two cars before their collision is

m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j

= (19,600 i + 46,000 j) kg•m/s

Their momentum after the collision is

(1400 kg + 2300 kg) v = (3700 kg) v

where v is the velocity vector of the wreckage.

By conservation of momentum,

(19,600 i + 46,000 j) kg•m/s = (3700 kg) v

Let a and b be the horizontal and vertical components of v, respectively. Then

19,600 kg•m/s = (3700 kg) a ⇒ a ≈ 5.2973 m/s ≈ 5.3 m/s

46,000 kg•m/s = (3700 kg) b ⇒ b ≈ 12.4324 m/s ≈ 12 m/s

so that the final speed of the wreckage is

||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s

3 0

2 years ago

How can you increase the pressure on an object?

Cerrena [4.2K]

Increase the force on the object

8 0

3 years ago