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To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

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Answer: A or B

Explanation: I’m guessing that they even each other out depending on the incline, gravity will help keeping it in place

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

Answer:

Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.

The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.

Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.

A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.

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Answer:

hey mate......looks like the question is incomplete

Answer:

1.43m

Explanation:

Given data

m1= 50kg

m2=70kg

We are told that Sally m1= 50 sat at one end which is 2m from the center

Hence, the summation of clockwise moment = summation of anticlockwise moment

See the attached image for your reference

50*2= 70*x

100= 70x

x= 100/70

x=1.43m

Hence the mass m2 will be at the 1.43m mark for the net torque to be zero

Answer:

The answer is D.250N.

Explanation:

Answer:

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

[tex]2.898\times 10^{-10}\ \text{m}[/tex] x-ray region

Explanation:

T = Temperature

b = Constant of proportionality = [tex]2.898\times 10^{-3}\ \text{m K}[/tex]

[tex]\lambda[/tex] = Wavelength

[tex]T=10^4\ \text{K}[/tex]

From Wein's law we have

[tex]\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}[/tex]

The wavelength of the radiation will be [tex]2.898\times 10^{-7}\ \text{m}[/tex] and it is in the ultraviolet region.

[tex]T=10^7\ \text{K}[/tex]

[tex]\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}[/tex]

The wavelength of the radiation will be [tex]2.898\times 10^{-10}\ \text{m}[/tex] and it is in the x-ray region.

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

Answer:

T

beacuse:

Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).

Answer:

Is this your question? Also I think work done is a scalar quantity.

Explanation:

Answer:

t = 9.52 s

Explanation:

This is an oscillatory motion exercise, in which the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

        F_e - W = 0

        F_e = W

        k x = m g

        k = [tex]\frac{m g}{x}[/tex]

        k = 0.545 9.8 /0.0356

        k = 150 N / m

now the angular velocity is related to the period

          W = 2π / T

we substitute

          4π² T² = k /m

          T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]

we substitute

           T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]

            T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

            t = 4 T

            t = 4 2.38

            t = 9.52 s

Answer:

Explanation:

this is like rubbing a balloon on your head to make your hair stand up.  Do that to the can.    The balloon is filled , ofc,  and then just rub the balloon on the can.  This will charge the can with static electricity.   :P

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

[tex] F = ma [/tex]

[tex] a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2} [/tex]

Now, we can calculate the final speed of the crate at the end of 10.0 m:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{1} [/tex]                  

[tex] v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s [/tex]    

For the next 10.5 meters we have frictional force:

[tex] F - F_{\mu} = ma [/tex]

[tex] F - \mu mg = ma [/tex]

So, the acceleration is:

[tex] a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2} [/tex]

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

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

[tex] v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s [/tex]  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

Answer:

    w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

Explanation:

To solve this exercise, let's use the concept of conservation of the angular number.

We create a system formed by the two discs, in this case the forces last the shock are internal

initial instant .. just before shock

           L₀ = I₀ w₀ + I₁ w₁

instnte final. Right after crash

           L_f = (I₀ + I1) w

angular momentum is conserved

           I₀ w₀ + I₁ w₁ = (I₀ + I₁) w

           w = I₀ w₀ + I₁ w₁ / Io + I1

The moment of inertia of a disk with an axis passing through its thermometric center

          I₀ = ½ m² r₀²

          I₁ = ½ m₁ r₁²

         we substitute

          I₀ = ½ 2.0 0.70²

          I₀ = 0.49 kg m

          I₁ = ½ 2.0 0.5²

          I₁ = 0.25

₁  

let's reduce the magnitudes the SI system

          w₀ = -50 rev / (2pi rad / 1rev) = -314.15 rad / s

          w₁ = 70 rev (2pi rad / 1rev) = 439.82 rad / s

we will assume that the counterclockwise turns are positive

           w = -0.49 314.15 + 0.25 439.82 / (0.49 + 0.25)

            w = (- 4.696 + 1.0995) 102) / 0.74

            w = -197.75 + 0.25

            w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

Answer:

Total energy = 1000J

KE = 500J

PE = 500J

Explanation:

As you may know, the equation for gravitational potential energy is mgh (weight x height)

If the skateboard is halfway down, that means it is at half the height. As the skateboard speeds up (as it goes downward), the potential energy becomes kinetic energy. Since it has 500J of kinetic energy at half way down, it means it had double that amount of Potential energy at the top (1000J). Since half of that became kinetic energy, there is only 500J of PE left.

Total energy = KE + PE = 1000J

KE = 500J

PE = 500J

Answer:

F = 812.25 N

Explanation:

Given (convert to SI units):

m = 57 g = 0.057 kg

[tex]v_{1}[/tex] = -25 [tex]\frac{m}{s}[/tex]

[tex]v_{2}[/tex] = 32 [tex]\frac{m}{s}[/tex]

t = 4 ms = 0.004 s

Find acceleration:

a = [tex]\frac{v_{2}-v_{1} }{t}[/tex] = [tex]\frac{32-(-25)}{0.004}[/tex] = 14250

Find force:

F = ma = 0.057(14250) = 812.25 N

Answer:

where a picture

Explanation:

a saan Ang picture

Answer:Unlike fixed joints or cartilaginous joints, where the bones are connected by either connective tissue or cartilage, the bones in synovial joints are not directly joined by anything, which allows for a much greater range of motion.

Explanation: I got a 100% on my test

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

The velocity of each electron at the corners of the square is 15.92 m/s.

The given parameters;

The diagonal length of the square is calculated as;

[tex]d^2 = 0.8^2 + 0.8^2\\\\d = \sqrt{0.8^2 + 0.8^2} \\\\d = 1.13 \ m[/tex]

The distance of each corner charge and the middle charge is calculated as;

[tex]r = \frac{1.13}{2} \\\\r = 0.565 \ m[/tex]

The force between each corner charge and the middle charge is calculated as;

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

The centripetal force on each charge moving around the square is calculated as;

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

solve the forces together;

[tex]\frac{mv^2}{r} = \frac{kq^2}{r} \\\\v^2 = \frac{kq^2}{m} \\\\v = \sqrt{ \frac{kq^2}{m} } \\\\v = \sqrt{ \frac{(9\times 10^9) \times (1.602\times 10^{-19})^2}{9.11 \times 10^{-31}} } \\\\v = 15.92 \ m/s[/tex]

Thus, the velocity of each electron at the corners of the square is 15.92 m/s.

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Answer:

Explanation:

From the given information:

Let the first weight be [tex]m_ 1[/tex] = 80 kg

The weight of the buddy be [tex]m_2[/tex] = 120 kg

The weight of  Bubba be [tex]m_3[/tex] = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

[tex]x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2[/tex]

The length of the boat be [tex]x_2[/tex] = 4 m

∴

We can find the center of mass of the system by using the formula:

[tex]X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}[/tex]

Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.

Given the following data:

Mass A = 5 lb to kg = 2.27 kg.

Mass B = 20 lb to kg = 9.02 kg.

Force = 3.6 lb to N = 16.02 Newton.

In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.

For cylinder A:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]

For cylinder B:

[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]

For the belt, we have

[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]

Also, we would determine the angular acceleration of cylinder B:

[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]

Next, we would calculate the forces acting on the cylinders:

[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]

[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]

Next, we would determine the force of static friction:

[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]

From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.

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