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Tom [10]
3 years ago
14

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

s the best approximation of the box’s acceleration?
120 m/s2
20 m/s2
1 m/s2
9 m/s2
Physics
1 answer:
wolverine [178]3 years ago
8 0

Answer:

1 m/s^{2}

Explanation:

The Net Force refers to the sum of all forces acting on an object. Hence,

Net Force of box

= 20 + (-9.0)       [It is -9.0 as both forces are in different directions]

= 11.0 N

We also need to know that the net force on an object is equal to the product of the mass of the object and the acceleration of the object. Hence,

F_{net} = ma

11.0 = 12.0a

a = 1 m/s^{2} (nearest whole number)

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ILL MARK BRAINLIEST
77julia77 [94]

Answer: B

Longitudinal wave

Explanation:

Transverse waves have crests and troughs

Longitudinal waves have compressions and rarefactions. A compression is where the density of the wave medium is highest. While a rarefaction is where the density of the wave medium is lowest.

Since sound wave is a longitudinal wave. And longitudinal wave exists apart from sound, we can therefore conclude that it's a longitudinal wave in spring.

8 0
3 years ago
How many protons are in the radioactive isotope 40/19K?
trasher [3.6K]
First we have to establish that the number of protons is equivalent to the atomic number of element. Here I am assuming that you are referring to Potassium (K) - 40. Potassium, stable or unstable has 19 protons.
8 0
3 years ago
a ladybug sits at the outer edge of a merry-go-round and a gentleman bug sits halfway between her and the axis of rotation. The
forsale [732]
The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)
6 0
3 years ago
9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
While participating in a blood drive at school, Keona learns that blood has a density of 1.06 g/mL. She donates one pint of bloo
Sergio [31]

Answer:

m≈501.57 g

Explanation:

The density formula is:

d=m/v

Let’s rearrange the formula for m. m is being divided by v. The inverse of division is multiplication, so multiply both aides by v.

d*v= m/v*v

d*v=m

The mass can be found by multiply the density and the volume.

m=d*v

The density is 1.06 grams per milliliter and the volume is 473.176 milliliters.

d= 1.06 g/mL

v= 473.176 mL

Substitute the values into the formula.

m= 1.06 g/mL * 473.176 mL

Multiply. When multiplying, the mL will cancel out.

m= 501.56656 g

Let’s round to the nearest hundredth. The 6 in the thousandth place tells us to round the 6 to a 7 in the hundredth place.

m ≈501.57 g

The mass is about 501.57 grams.

7 0
3 years ago
Read 2 more answers
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