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natima [27]
3 years ago
15

If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder

Physics
1 answer:
Natali [406]3 years ago
6 0
The baseball because it has a greater mass and would go farther because it don't have holes for the air to go through and slow it down
You might be interested in
In all measurements and calculations, magnitude (number) and units are always included.
san4es73 [151]

the equation of the line allows us to find the answer is

            y = -27.8 t + 97.4

The equation of a line in a linear relationship between two variables, its general expression is

         y = A x + B

in this case the slope is the quantity that the independent variable in this case A = -27.8 m / s

The cut-off point that is the value of the dependent variable for x = is b = 97.4 m

In this case we see that the slope has a unit of [m / s] and the dependent variable is a unit of length, therefore the independent variable must have a unit of time [s] so that the entire equation is in units of length

          y = -27.8 t + 97.4

         [m] = [m / s] [s] + [m]

         [m] = [m]

The other two magnitudes with are necessary to write the equation r is the mean square root and gives an idea that the values ​​also fit the line, the best value is 1

In conclusion, the equation of the line allows us to find the answer is

            y = -27.8 t + 97.4

learn more about the equation inear here:

brainly.com/question/22851869

8 0
3 years ago
На рычаге размещены два противовеса таким образом, что рычаг находится в состоянии равновесия. Вес расположенного слева противов
prisoha [69]

Answer:

Вес противовеса, или сила нагрузки, составляет затем 100 000 фунтов-футов, разделенных на 20 футов, или 5000 фунтов.

Explanation:

8 0
3 years ago
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
3 years ago
A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes
mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

d_1 = v* t

d_1 = 20 * 0.8

d_1 = 16 m

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2*(-7)*d_2

d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

3 0
3 years ago
A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

6 0
3 years ago
Read 2 more answers
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