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hammer [34]
3 years ago
5

At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himsel

f. Mason pulled the bat with
force of 15 newtons (N) while Alfredo pulled with a force of 20 newtons (N). Why did Alfredo end up with the bat?

A because the force was 5 N in Mason's direction

B. O because the net force was 5 N in Alfredo's direction

c. O because the net force was 15 N in Mason's direction

D.O because the net force was 20 N in Alfredo's direction
Physics
1 answer:
Illusion [34]3 years ago
7 0

Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

Fₙ = 20 – 15

Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

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A flat polished area is turned through 15 degrees. If the angle of incidence is 20 degrees, what is the angle between the incide
Ilya [14]

Answer:

<h2>50°</h2>

Explanation:

Angle of rotation of the flat polished surface \theta = 15°

angle of incidence i = 20°

Since the polished surface is turned at an angle of 15°, the angle of reflection

r = 2\theta (Note that the angle of rotation only have effect of the angle of reflection)

r = 2*15 = 30°

The angle between the reflected ray and the incident ray will be equal to the sum of the angle of incidence and the angle of reflection i.e i+r

The angle between the reflected ray and the incident ray = 20°+ 30° = 50°

7 0
4 years ago
A ball dropped from a height of 50 meters, Determine the speed of the ball after 3 seconds
Alex17521 [72]

16.7 m/s

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8 0
4 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
It has been theorized that all of the modern continents on Earth were once part of a single continent. What scientific evidence
HACTEHA [7]

Answer:

B

Explanation:

It is not just one of these answers, in fact they all make sense. But the one I learned in school was that when the continents were still one super continent- Pangea, animals fossilized wear the borders of modern day continents were. This later on scientists found half of an animal fossil on the edge of one continent and the other half on the edge of a different continent.

hope this helps

~Lalagrl

4 0
4 years ago
Read 2 more answers
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