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Katena32 [7]
3 years ago
9

A baseball is moving at a speed of 2.2\,\dfrac{\text{m}}{\text{s}}2.2 s m ​ 2, point, 2, space, start fraction, m, divided by, s

, end fraction when it strikes the catcher's glove. The padding of the glove is compressed by 2.4\,\text{mm}2.4mm2, point, 4, space, m, m before the ball comes to a stop. We want to find the average acceleration of the baseball while it is compressing the glove.
Physics
1 answer:
Reika [66]3 years ago
5 0

Answer:

 a = 1,008 10⁻³ m / s²

Explanation:

For this exercise, let's use the kinematic relations of accelerated motion

           v² = v₀² - 2 a x

The negative sign is because the acceleration is opposite to the speed, the final speed is zero

          0 = v₀² - 2 a x

          a = v₀² / 2x

   Let's reduce the magnitudes to the SI system

          x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m

Let's calculate

         a = 2.2²/2 2.4 10⁻³

        a = 1,008 10⁻³ m / s²

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A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The
almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

             Initial momentum = Final momentum

                                       Pi = Pf

                  M*vs_e = m*vc_e + 4m*vm_e

Where,

                  M = m + 4m = 5m

                  vc_e = Velocity of command relative to earth

                  vm_e = Velocity of motor relative to earth  

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

                  vm_e =  v_c/m + vc_e        

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

                  5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

                  5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

                  5m*vs_e -  4m*(v_c/m) = 5m*vc_e

                   vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

                  vc_e = 3760 - 0.8*(90)

                  vc_e = 3,688 km/h

5 0
3 years ago
Read 2 more answers
Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values
Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and  4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and  1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and  2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and  2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge   Right Charge Resulting force(N)

1μC                     4μC                  90 N

4μC                   1μC                    90 N

2μC                  2μC                    90 N

1μC                    2μC                   45 N

1μC                  8μC                    180 N

2μC                  8μC                  360 N

5 0
3 years ago
The density of a material in CGS system of units is 4g cm-³. In a system of units in which unit of length is 10 cm and unit of m
butalik [34]

\sf\underline{Solution:}

Here , the density of the material is 4g cm³ but it is not given in CGS system.

$\sf{As\:we\:know\:that:}$

$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$

$\space$

\sf{Now,according \: to \:the\:question:}

$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$

$\space$

$\sf{It\:is\:given\:that:}$

In the system of units the mass is 100gram.

$\space$

Hence,

$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$

$\space$

In the system of units,the length is 10cm.

Henceforth,

$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$

$\space$

<u>☆</u><u> </u><u>Substitute</u><u> </u><u>the</u><u> </u><u>required</u><u> </u><u>values</u><u> </u><u>in</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>formula</u><u>-</u>

$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$

$\space$

$\sf\underline\bold{Density\:of\:the\:material:}$

= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$

$\space$

$\sf\underline\bold\blue{=40\:units}$

$\sf\small{Therefore,option\:2nd\:is\:correct!}$

_______________________________

6 0
3 years ago
Which of these best describes how an appropriate star chart is selected to locate objects in the sky?
Sergio [31]
The layout of the stars in the sky is determined by the date, time of night, and your location (mainly latitude). So to pick the best star chart, you should go with the one that's closest to the present date and your location, then make allowance for what time it is. Everything in the sky moves about a degree every 4 minutes.
6 0
3 years ago
Read 2 more answers
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon
Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

F_{branch} + F_{gravity}=F_{centripetal}

force of gravity = - mg

F_{branch}=F_{centripetal}-F_{gravity}

                        = \dfrac{mv^2}{r} + mg

                        = m(\dfrac{3.4^2}{0.6} +9.8)

                        =9 x 29.067

                        = 261.6 N

upward force acting = 261.6 N

7 0
3 years ago
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