In a little league baseball game, the 145 grams ball enters the strike zone with a speed of 11.0 meters per second. The batter hits the ball and it leaves his bat with a speed of 25.0 meters per second in exactly the opposite direction. If the bat is in contact with the ball for 1.0 m/s, what is the magnitude of the average force exerted by the bat on the ball?​

Answers

Answer 1

Answer:

Force = 5.22 N

Explanation:

According to Newton's Second Law of motion:

[tex]Force = Rate\ of\ Change\ of\ Momentum\\\\Force = \frac{mv_f-mv_i}{t}\\[/tex]

where,

m = mass of ball = 145 g = 0.145 kg

vf = final speed of ball after hit = 25 m/s

vi = initial speed of ball before hit = -  11 m/s (negative sign due to opposite direction)

t = time of contact = 1 s

Therefore,

[tex]Force = \frac{(0.145\ kg)(25\ m/s)-(0.145\ kg)(-11\ m/s)}{1\ s} \\\\[/tex]

Force = 5.22 N


Related Questions

which statement can best describe the energy transformations that occur when a pendulum swings back and forth?
a. gravitational energy is converted to spring energy and back.
b. gravitational energy is converted to kinetic energy and back
c. kinetic energy is converted to spring energy and back
d. none of the above.​

Answers

Answer:

I think it is C, I'm not for sure.

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Answers

Answer:

the force of attraction between the two charges is 4.2 x 10 N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

[tex]F = \frac{Kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]

Therefore, the force of attraction between the two charges is 4.2 x 10 N.

A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.​

Answers

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg

and the incline is µ1 = 0.20 and that between the block of mass 4.0 kg and the

incline is µ2 = 0.30. Find the acceleration of 2.0 kg block. ( g = 10m/s^2).​

Answers

Answer:

The acceleration of 2.0 kg block is 2.7 m/s²

Explanation:

Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:

Force down the plane on the system

= (4 + 2) g sin30°

= (6)(10)(½)

= 30N

Force up the plane on the system

= µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°

= (2µ₁ + 4µ₂) g cos30°

= (2 × 0.2 + 4 × 0.3)(10)(√3/2)

≈ 13.76 N

∴ Net force down the plane is F

F = 30 - 13.76

F = 16.24 N

∴Acceleration of both the blocks down the

plane will b a

a = F ÷ (4 + 2)

a = 16.24 ÷ 6

a = 2.7 m/s²

Thus, The acceleration of 2.0 kg block is

2.7 m/s²

-TheUnknownScientist

Explanation:

2.7m/s2

I hope its helpful

What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?

Answers

Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!  

bit.[tex]^{}[/tex]ly/3a8Nt8n


An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.

Answers

Answer:

soluble soluble soluble soluble

Explanation:

solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci

The nucleus of a certain type of uranium atom contains
92 protons and 143 neutrons. What is the total charge of
the nucleus?

Answers

Answer:

charge = electrons + protons

=92+92

=184

Which is a property of bases?

A.
highly metal reactive

B.
sour to the taste

C.
slippery feel

D.
low pH

Answers

Answer:

C. Slippery feel

Explanation:

Slippery feel would be the correct answer (:

A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs . Suppose a technician holds his hand near the device so that the axis of his 2.0-cm-diameter wedding band is parallel to the field.
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
ε = ___________
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
I = ____________

Answers

Answer:

A. 3.9 V B. 1.9 fA

Explanation:

Part A

What emf is induced in the ring as the field changes?

Express your answer to two significant figures and include the appropriate units.

The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time

ε = ΔΦ/Δt

ε = ΔABcos0°/Δt

ε = AΔB/Δt

A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T.  ΔB = B₁ - B₀  = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.

So, ε = AΔB/Δt

ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s

ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s

ε = 3.926 V

ε ≅ 3.9 V

Part B

If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.

Express your answer to two significant figures and include the appropriate units.

Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².

So, i =  ε/R

=  ε/ρl/A

= εA/ρl

= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)

= ‭15.704‬ × 10⁻⁶ V-m²/(82.9356‬ × 10⁸ Ωm²

= 0.1894 × 10⁻¹⁴ A

= 1.894 × 10⁻¹⁵ A

≅ 1.9 fA

A 6kg object undergoes an acceleration of 2m/s, what is the magnitude of the resultant acting on it . If this same force is applied to a 4kg object, what acceleration is produced

Answers

Answer:

[tex]12\; \rm N[/tex].

[tex]3\; \rm m\cdot s^{-2}[/tex].

Explanation:

By Newton's Second Law, the acceleration of an object is proportional to the size of the resultant force on it, and inversely proportional to the mass of this object.

[tex]\displaystyle \text{acceleration} = \frac{\text{resultant force}}{\text{mass}}[/tex].

Rearrange this equation for the resultant force on the object:

[tex]\text{resultant force} = \text{acceleration} \cdot \text{mass}[/tex].

For the [tex]6\; \rm kg[/tex] object in this question:

[tex]\begin{aligned} F &= m \cdot a \\ &= 6\; \rm kg \times 2\; \rm m \cdot s^{-2} \\ &=12\; \rm N\end{aligned}[/tex].

When the resultant force on the [tex]4\; \rm kg[/tex] object is also [tex]12\; \rm N[/tex], the acceleration of that object would be:

[tex]\begin{aligned} a &= \frac{F}{m} \\ &= \frac{12\; \rm N}{4\; \rm kg} = 3\; \rm m \cdot s^{-2}\end{aligned}[/tex].

A radioactive material produces 1160 decays per minute at one time, and 4.0 h later produces 170 decays per minute. whats the half life

Answers

Answer:

Half life is 3.23 hours

Explanation:

Given

Decay rate at starting = 1160 decays per minute

Decay rate after 4 hours = 170 decays per minute

As we know know

[tex]N = N_0 *e ^{\Lambda *T}[/tex]

Substituting the given values, we get -

[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]

Also

[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]

Substituting the given values we get -

[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours

In Part 5.2.3 of the experiment, you will measure the index of refraction of yellow light using Lab Manual Equation 5.2. Suppose the minimum angle of deviation is 18 degrees. What is the index of refraction

Answers

Answer:

The answer is "1.26".

Explanation:

[tex]D=18^{\circ}[/tex]

The refractive index is:

[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]

       [tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]

A man pushes a 10-kg block 10 m, along a rough, horizontal
surface with a 40-N force directed 37° below the horizontal. If
the coefficient of kinetic friction is 0.2, calculate the total
work done on the block

Answers

Answer:

hope you can understand

A car is travelling at 27m/s and decelerates at a=5m/s2 for a distance of 10m. Calculate its final velocity. (Hint does deceleration imply that the acceleration is positive or negative?)[

Answers

Answer:

use the formula to calculate acceleration and you'll get the answers

An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byemitting a pulse of ultrasound into air and then measuring the timeinterval for an echo to return from a reflecting surface whosedistance away is to be measured. The distance is displayed as adigital read-out. A tape measure emits a pulse of ultrasound with afrequency of 25.0 MHz.
(a) What is the distance to an object fromwhich the echo pulse returns after 24ms when the air temperature is 26°C?
(b) What should be the duration of the emitted pulse if it is toinclude 10 cycles of the ultrasonic wave?
(c) What is the spatial length of such a pulse?

Answers

Answer:  

a) 1m

b) 2μs

c) 3mm

Explanation:

1. Objects become electrically charged as a result of the transfer of

Answers

Answer:

Electron

Explanation:

An object can become electrically charged when it gains or loses an electron. Because an electron is negatively charged, when an object gains an electron it becomes negatively charged. Also, when it gives up an electron, it becomes positively charged. This positive charge is because the atom has one proton more than electron. In a neutral atom, the number of the proton is equal to the number of the electron. An electron is negatively charged, and a proton is positively charged.

While visiting Planet Physics, you toss a rock straight up at 15 m/s and catch it 2.7 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet's radius every 250 min .
Part A What is the mass of Planet Physics?
Part B What is the radius of Planet Physics?

Answers

Answer:

R = 7.915 10⁶ m,  M = 1.04 10³⁵ kg

Explanation:

Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations

           v = v₀ - g t

the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls

          v = -v₀

         -v₀ = v₀ - g t

          g = 2v₀ / t

          g = 2 15 / 2.7

          g = 11.11 m / s²

I now write the law of universal gravitation and Newton's second law

          F = m a

          G m M / R² = m a

         a = g

          g = G M / R²

           

Now let's work with the cruiser in orbit

         F = ma

acceleration is centripetal

         a = v² / r

         G m M / r² = m v² / r                        (1)

the distance from the center of the planet is

        r = R + h

        r = R + R = 2R

we substitute in 1

        G M / 4R² = v² / 2R

        G M / 2R = v²

The modulus of the velocity in a circular orbit is

         v = d / T

the distance is that of the circle

          d = 2π r

          v = 2π 2R / T

          v = 4π R / T

          G M / 2R = 16pi² R² / T²

          T² = 32 pi² R³ / GM

let's write the equations

             g = G M / R² (2)

             T² = 32 pi² R³ / GM

we have two equations and two unknowns, so it can be solved

let's clear the most on the planet and equalize

             g R² / G = 32 pi² R³ / GT²

              g T² = 32 pi² R

             R = g T² / 32 pi²

let's reduce the period to SI units

           T = 250 min (60 s / 1 min) = 1.5 104 s

let's calculate

             R = 11.11 (1.5 10⁴) ² / 32 π²

           R = 7.915 10⁶ m

from equation 2 we can find the mass of the planet

             M = g R² / G

             M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹

             M = 1.04 10³⁵ kg

Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the
end the compass points away from as "S" (south), and the other end that the compass points toward as "N" (north).
Record these poles in Figure 1.
Continue
Intro

Answers

Answer:

the red pointer on the magnet ( grey region) : points towards north

red pointer outside the magnet ( white region) is pointing towards south

Explanation:

please see the attached image

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface

Answers

This question is incomplete, the complete question is;

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer:

the required wavelength of light is 161.9 nm

Explanation:

Given the data in the question;

Let us represent work function of the metal by W.

Now, using Einstein photoelectric effect equation;

[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]

hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1

we solve for W

W = hc/λ - [tex]K_{max[/tex]

given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J

we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.

so we substitute

W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] -  3.28 × 10⁻¹⁹

W =  8.99457 × 10⁻¹⁹ -   3.28 × 10⁻¹⁹

W = 5.71457 × 10⁻¹⁹ J

Now, to determine λ for which maximum kinetic energy is double

so;

[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J

from from equation 1

we solve for λ'

λ' = hc / W + [tex]K'_{max[/tex]

we substitute

λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))

λ' = 1.9878 × 10⁻²⁴ /  1.227457 × 10⁻¹⁸

λ' =  1.619 × 10⁻⁷ m

λ' =  161.9 nm

Therefore, the required wavelength of light is 161.9 nm

The unit of work done is called derived unit why​

Answers

Energy can be measured as work. We can write energy = force x distance. Thus SI derived unit of energy has the units of newtons x meter or kg m2/s2.

What are the similarities between a resultant force equilibrant force?

Answers

Answer:

Explanation:

Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.

Drag each statement to the correct location on the table.
Match the characteristics with the states of matter.
does not have
a definite shape
or volume
has definite volume
but does not have a
definite shape
has a definite shape
and volume
changes to liquid
on heating
changes to liquid
on cooling
changes to solid
on cooling
Solid
Liquid
Gas
mentum. All rights reserved.

Answers

Answer:

Solid:

- has definite shape and volume.

- change to liquid on heating.

Liquid:

- has definite volume but does not have definite shape .

- changes to solid on cooling.

Gas :

- does not have definite shape or volume.

- change to liquid on cooling

A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed v0 collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. What applications of the equation for the conservation of momentum represent the initial and final momentum of the system for a completely inelastic collision between the blocks?

Answers

Solution :

In the question, it is given that the collision is inelastic and the blocks stick together.

In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.

The linear momentum is given by :

[tex]$\vec p = m \vec v$[/tex] (mass x velocity)

So according to the conservation of linear momentum,

[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]

Let the velocity after the collision is [tex]$v_F$[/tex]

[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]

Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]

[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]

∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.

and  [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.

What happens when Earth rotates on its axis and how long does it take

Answers

Answer:

You get Day and Night

It takes 24 hour

Answer:

Explanation:

The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.

To enhance heat rejection from a spacecraft, an engineer proposes to attach an array of rectangular fins to the outer surface of the spacecraft and to coat all surfaces with a material that approximates blackbody behavior. Consider the U-shaped region between adjoining fins and subdivide the surface into components associated with the base (1) and the side (2). Obtain an expression for the rate per unit length at which radiation is transferred from the surfaces to deep space, which may be approximated as a blackbody at absolute zero temperature. The fins and the base maybe assumed to be isothermal at a temperature T. Comment on your result. Does the engineer's proposal have merit

Answers

Answer:

Attached below is the required diagram related to the question

answer :

q'3 = WбT^4

engineer's proposal has merit

Explanation:

Let : A'3 represent the deep space

      A'1 represent the surface area , F13 and  F23  represent the view factors

      T1 , T2, T3 ; represent temperatures

      q'3 represent net rate of heat radiation

 Derive the expression for the rate per unit length at which radiation is transferred from the surfaces to deep space

derived expression ; q'3 = WбT^4

attached below is a detailed solution

Given that The emission is proportional to the area of the opening and the surfaces ( 1 and 2 ) have the same temperature hence this problem can be treated as a two surface enclosure. hence the engineer's proposal have merit .

attached below is a prove ( b )

If you are in a car that is traveling at 60 mph and a balanced force is applied to the car what will happen to the motion of the car? /gen

Answers

Answer:

The car will stop moving

Explanation:

If a balanced force is applied to the car in motion, the car will stop accelerating and remain stationary because all the forces acting on the car are equal.

Also, you can say, since a balanced force is applied, the net force or resultant force on the car is zero. According to Newton's second law of motion, an object will move in the direction of the applied force. When the resultant force on the object is zero, it means the object will not move.

According to the Traditional Square of Opposition: If "All S are P" is true, then is "Some S are P" true or false?

Answers

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the concept of Logic.

Since in the above statement it is given that,

All S are P ==> True,

then obviously Some S are also P always, hence it is true.

Answer is True.

12. By convention (agreement of the scientific community for consistency)
magnetic field lines...

A. always start on the north pole and terminate (end) on the South Pole

B. start at infinity and point toward each pole

C. start at each pole and go outward

D. always start on the south pole and terminate (end) on the north pole.

Answers

Answer:

. always start on the north pole and terminate (end) on the South Pole

Explanation:

Which of these is NOT an inherited trait of the plant?

Answers

Explanation:

I think you forgot to add the other part!

Answer:

uhm what plant...

Explanation:

Which of these will be the correct relationship between work input and work output?
A) Work input = Work output + Work against friction
B) Work input = Work output – Work against friction
C) Work input = Work output * Work against friction
D) Work input = Work output / Work against friction

Answers

Answer:

Work input = Work output * Work against friction is your answer so C

Explanation:

I hope this helps you :)

Answer:

A) Work input = Work output + Work against friction

Explanation:

Other Questions
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