I need help!!!!!!!!!!!pleaseeeeeee

Answer:

(a) The change in momentum is 12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

Explanation:

Given that,

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]

(b) Time for collision, t = 0.012 s

Now the force can be calculated as follows :

[tex]F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N[/tex]

Hence, this is the required solution.

Answer:

a. = 12.04 kg*m/s

b. = 1,003.3N

Explanation:

The answer above is correct.

Answer:

The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.

Answer:

A and C

Explanation:

Both have mass and are in motion

Answer:

1) the required horizontal force F is 1095.6 N

2) W = 0 J { work done by rope will be 0 since tension perpendicular }

3) work is done by the worker is 1029.4 J

Explanation:

Given that;

mass of bag m = 125 kg

length of rope [tex]l[/tex] = 3.3 m

displacement of bag d = 2.2 m

1) What horizontal force is necessary to hold the bag in the new position?

from the figure below; ( triangle )

SOH CAH TOA

sin = opp / hyp

sin[tex]\theta[/tex] = d / [tex]l[/tex]

sin[tex]\theta[/tex] = 2.2/ 3.3

sin[tex]\theta[/tex] = 0.6666

[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )

[tex]\theta[/tex]  = 41.81°

Now, tension in the string is resolved into components as illustrated in the image below;

Tsin[tex]\theta[/tex] = F  

Tcos[tex]\theta[/tex] = mg

so

Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg

sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg

we know that; tangent = sine/cosine

so

tan[tex]\theta[/tex] = F / mg

F = mg tan[tex]\theta[/tex]

we substitute

Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )

F = 1225 × 0.8944

F = 1095.6 N

Therefore, the required horizontal force F is 1095.6 N

2)  As the bag is moved to this position, how much work is done by the rope?

Tension in the rope and displacement of mass are perpendicular,

so, work done will be;

W = Tdcos90°

W = Td × 0

W = 0 J { work done by rope will be 0 since tension perpendicular }

3) As the bag is moved to this position, how much work is done by the worker

from the diagram in the image below;

SOH CAH TOA

cos = adj / hyp

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]

we substitute

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]  = 1 - h/[tex]l[/tex]

cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]

h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]

h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

now, work done by the worker against gravity will be;

W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

we substitute

W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )

W = 4042.5 × ( 1 - 0.745359 )

W = 4042.5 × 0.254641

W = 1029.4 J

Therefore,  work is done by the worker is 1029.4 J

Answer:

a)      N = 7.90 10³ N,  b) N = -1.04 10⁴ N

Explanation:

a) For this exercise we can use Newton's second law

             N -W = m (-a)

The relationship is centripetal, the negative sign of the acceleration is because it points towards the center of the circle

              a = v² / r

we substitute

             N = mg -m v² /r  

             N = m (g - v² /r)

let's calculate

v = 8.7 m / s

             N = 1240 (9.81 - 8.7²/22)

             N = 7.90 10³ N

b) v = 20 m / s

             N = 1240 (9.81 - 20²/22)

             N = -1.04 10⁴ N

Answer:

0.0s

Explanation:

I got it right in acellus

Answer: 0.0

Explanation:

Answer:

94.25 seconds

Explanation:

Solve for period (T) using: v=(2*pi*r)/T

rearrange: vT=2*pi*r

rearrange: T=(2*pi*r)/v

Plug in values.

T=(2*pi*150)/10

T=94.25 seconds

If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.

As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.

vT = 2 × π × r

T = (2 × π × r)/ v

T = (2 × π× 150)/10

T = 94.25 seconds

Thus, the time taken to complete the one lap would be 94.25 seconds.

To learn more about speed here, refer to the link given below ;

https://brainly.com/question/7359669

#SPJ2

Answer:

λ = 5.434 x 10⁻⁷ m = 543.4 nm

Explanation:

To solve this problem we can use the formula provided by Young's Double Slit experiment:

[tex]\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta xd}{L}[/tex]

where,

λ = wavelength of light = ?

Δx = distance between adjacent bright fringes = 1.6 cm = 0.016 m

d = slit separation = 0.18 mm = 0.00018 m

L = Distance between slits and screen = 5.3 m

Therefore,

[tex]\lambda = \frac{(0.016\ m)(0.00018\ m)}{5.3\ m}[/tex]

λ = 5.434 x 10⁻⁷ m = 543.4 nm

Answer:

The sun and the stars

Explanation:

I hope this helps!

Answer:

Northstar & North Pole

Explanation:

yyggggggggggggggg

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

Spring constant in N / m = 6,000

Answer:

[tex]a=0.13m/s^2[/tex]

Explanation:

From the question we are told that

Mass of first team man [tex]m_1=64kg[/tex]

Force of man first team man [tex]F_1=1350[/tex]

Mass of second team man [tex]m_2=69kg[/tex]

Force of man second team man [tex]F_2=1367N[/tex]

Generally the equation for net force F_n is mathematically given by

[tex]F_n=9(m_1+m_2)a[/tex]

[tex]9(m_1+m_2)a=9(f_2-f_1)[/tex]

[tex]9(64+69)a=9(1367-1350)[/tex]

[tex]a=\frac{9(1367-1350)}{9(64+69)}[/tex]

[tex]a=0.127819m/s^2[/tex]

Therefore the acceleration is given by

[tex]a=0.13m/s^2[/tex]

Answer:

The period is more, but not doubled.

Explanation:

Recall that the period of a mass on a spring is T=2πmk.

Answer:

Wavelength = 1.04 meters

Explanation:

Given the following data;

Speed = 509m/s

Frequency = 491Hz

To find the wavelength;

Wavelength = speed/frequency

Wavelength = 509/491

Wavelength = 1.04 meters

Therefore, the guitar player should shorten the length of the string to 1.04 meters.

Answer: 0.52

Explanation:

Answer:

Materials that can lose or gain electrons in this way are called triboelectric, according to Northwestern University. One common example of this would be shuffling your feet across carpet, particularly in low humidity which makes the air less conductive and increases the effect.

Explanation:

Hope that helps :)

Answer:

Newton's theory identified mass as the factor that causes gravity. On the other hand, Einstein's theory identified the curvature of space-time as the factor that causes gravity.

Answer:

Hey mate...

Explanation:

This is ur answer....

In the 17th century Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's interpretation was that these objects do not fall. According to Einstein, these objects and Earth just freely move in a curved spacetime and this curvature is induced by mass and energy of these objects.

Hope it helps you,

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Answer:

A. 450 N

B. 3240 J

C. 0.77 KW

Explanation:

From the question given above, the following data were obtained:

Height of one step = 20 cm

Number of steps = 36

Mass of runner = 45 kg

Time taken = 4.2 s

Next, we shall convert 20 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

20 cm = 20 cm × 1 m /100 cm

20 cm = 0.2 m

Next, we shall determine the total height. This can be obtained as follow:

Height of one step = 0.2 m

Number of steps = 36

Total height =?

Total height = 36 × 0.2

Total height = 7.2 m

A. Determination of the runner's weight.

Mass of runner (m) = 45 kg

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

Weight (W) =?

W = m × g

W = 45 × 10

W = 450 N

B. Determination of the increase in the potential energy.

At the ground level, the potential energy (PE₁) is 0 J.

Next, we shall determine the potential energy at 7.2 m. This can be obtained as follow:

Mass of runner (m) = 45 kg

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

Total height (h) = 7.2 m

Potential energy at height 7.2 m (PE₂) = ?

PE₂ = mgh

PE₂ = 45 × 10 × 7.2

PE₂ = 3240 J

Final, we shall determine the increase in potential energy. This can be obtained as follow:

Potential energy at ground (PE₁) = 0 J

Potential energy at height 7.2 m (PE₂) = 3240 J

Increase in potential energy =?

Increase in potential energy = PE₂ – PE₁

Increase in potential energy = 3240 – 0

Increase in potential energy = 3240 J

C. Determination of the power.

Energy (E) = 3240 J

Time (t) = 4.2 s

Power (P) =?

P = E/t

P = 3240 / 4.2

P = 771.43 W

Finally, we shall convert 771.43 W to kilowatt (KW). This can be obtained as follow:

1000 W = 1 KW

Therefore,

771.43 W = 771.43 W × 1 KW / 1000 W

771.43 W = 0.77 KW

Therefore, her power is 0.77 KW

Answer: 6611.715 joules

Explanation:

Q = MxCxdeltaT = 6959.7 which is 100%

95% = 6611.715

Answer:

v = 3.74 m/s

Explanation:

Given that,

The acceleration of the object in circular path, a = 0.4 m/s²

The radius of the circle, r = 35 m

We need to find the velocity of the object. The acceleration of an object on the circular path is given by :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{0.4\times35}\\\\v=3.74\ m/s[/tex]

So, the velocity of the object is equal to 3.74 m/s.

Answer:

the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

Explanation:

To determine the advice to the runners, let's use the Farad equation to and

           fem = -N [tex]\frac{ d \phi}{dt}[/tex]  = -N [tex]\frac{ B A Cos \theta }{dt}[/tex]

how the runners are moving

                 fi = B l x

            fem = -N B l v

therefore the advice we can give are:

* the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

Answer:

A is the answer I think pls check it

Answer:

d = 700.4 m

Explanation:

Given that,

The speed of sound through air at 32.0°C is 350.2 m/s.

Reginald sees a bolt of lightning and hears the thunderclap 2.00s later.

We need to find how far from Reginald did the lightning strike. Let the distance be d. So,

Speed = distance/time

d = vt

So,

d = 350.2× 2

d = 700.4  m

So, the required distance is 700.4  m.

Answer:

2)

Explanation:

Answer:

1) v= 90km/h  d = 70 m,  2)  x₁ = v t_r,  x₁ = 6.25 m, 3) x₁=6.25 no change

4) x = 22 m

Explanation:

1) for the first part, you are asked to find the minimum safety distance with the vehicle in front

The internet is searched for the stopping distance for two typical speeds on the highway

v (km/ h)      v (m/s)      d (m)

90                  25           70

100                 27.78      84

the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement

              v = x / t_r

               x₁ = v t_r

the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus

therefore the safe distance is

              x_total = x₁ + d

2) The distance is the sum of the distance traveled in the reaction

              x₁ = v t_r

for v = 90 km / h

              x₁ = 25 0.25

              x₁ = 6.25 m

for v = 100 km / h

              x₁ = 27.78 0.25

              x₁ = 6.95 m

the total distance is

               x_total = x₁ + d

for v = 90 km / h

             x_total = 25 0.25 + 70

             x_total = 76.25 m

this is the distance until the cars stop and do not collide

3) the stopping distance of a truck is

   v = 90 km / h       d = 100 m

in this case we see that the braking distance is much higher,

the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change

4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s

for the car case at v = 90km / h = 25 m / s

according to this rule we must this to

           x = 25 / 4.47 = 5.6 cars

each modern car is about 4 m long so the distance is

           x = 22 m

we see that this distance is much greater than the reaction  distance so it does not make much sense

Answer:

25000J

Explanation:

Formula : Q = m×c×Δt

Q=Heat energy

m= mass

c=specific heat capacity

ΔT = change in temperature.

Q=M x C for this question.

Specific heat capacity = 2500 J/Kg

Mass = 10kg

2500 x 10 = 25000 J

Answer:

coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.

Explanation:

Viscosity is defined as the extent to which a fluid can resist flow when a force is applied to it.

Now, coefficient of viscosity is the term in which viscosity is calculated. It is basically the tangential frictional force acting per unit area which is divided by the velocity gradient as a result of streamline flow conditions.

Thus, coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load [tex]m_{LS}[/tex] = 10000 kg

mass of flat bed [tex]m_{FB}[/tex] = 20000 kg

initial speed of truck [tex]v_{0}[/tex] = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs [tex]F_{N}[/tex]     -------------let this be equation 1

where [tex]F_{N}[/tex] = normal force = mg

so

Fs,max = μs mg

ma[tex]_{max}[/tex] = μs mg

divide through by mass

a[tex]_{max}[/tex] = μs g    ---------- let this be equation 2

in equation 2, we substitute in our values

a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²

a[tex]_{max}[/tex] = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx

where [tex]v_{f}[/tex] is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² /  9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Answer: C. Charge and distance