All categories

3 years ago

You will get the most accurate resting heart rate if you take your pulse for ___ consecutive mornings and average the number

Physics

1 answer:

Setler79 [48]3 years ago

3 0

The answer will be 3

You might be interested in

A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its

larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

$F_n_e_t=ma$

Plug in what you know, and solve:

$18=m(12)\\m=1.5kg$

Find matching soluation:

C.) 1.5 kg

5 0

2 years ago

8) Find the X and Y component of 10degree vector that has 5N.

Deffense [45]

Answer:

Fx = 4.92 [N]

Fy = 0.868 [N]

Explanation:

Let's take the 10 degrees as a measure from the horizontal component to the vector.

Thus taking the components in the X & y axes respectively:

Fx = 5*cos(10) = 4.92 [N]

Fy = 5*sin(10) = 0.868 [N]

3 0

3 years ago

A hiker travels south along a straight line path for 1.5 hours with an average velocity of 0.75 km/hr, then travels south for 2.

ANTONII [103]

Distance travelled in south direction= 1.5hr*0.75km/hr= 1.125km
Distance travlled in north direction= 0.90*2.5=2.25
Net displacement = 2.25-1.125= 1.125 to the north

7 0

3 years ago

Which force is there between an ice skate and the ice

andrew-mc [135]

The answer is friction

7 0

3 years ago

Read 2 more answers

A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free e

Serhud [2]

Answer: The electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

Explanation:

Given: Current = 5.0 A

Area = $4.0 \times 10^{-6} m^{2}$

Density = 2.7 $g/cm^{3}$ , Molar mass = 27 g

The electron density is calculated as follows.

$n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\$

where,

$\rho$ = density

M = molar mass

$N_{A}$ = Avogadro's number

Substitute the values into above formula as follows.

$n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}$

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

8 0

3 years ago