How many elements are in the mixture pictured?<br> A. 7<br> B. 4<br> C. 3<br> D. 2

A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,

Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

8 0

3 years ago

In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it

Liono4ka [1.6K]

Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

= 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

5 0

3 years ago

A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?

Veronika [31]

Answer:

a

Explanation:

its correct

3 0

3 years ago

Convert 27,549 into scientific notation

dezoksy [38]

2.7549 x 10^4 is the answer I hope this helped u

7 0

3 years ago

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a

Marina86 [1]

Answer:

$I_{2}=2.39*10^{-5} W/m^{2}$

Explanation:

The definition of the intensity in terms of power is given by:

$I=\frac{P}{A}$

Where:

P is the power
A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is $A = 4\pi R^{2}$

To the first location we have:

$I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}$

and to the second location we have:

$I_{2}=\frac{P}{4\pi 78^{2}}$

Now, we can divide each intensity to find the second intensity.

$\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}$

$I_{2}=I_{1}* \frac{22^{2}}{78^{2}}$

$I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}$

$I_{2}=2.39*10^{-5} W/m^{2}$

I hope it helps you!

5 0

3 years ago

Read 2 more answers

How many elements are in the mixture pictured? A. 7 B. 4 C. 3 D. 2