The amount of work done on an object is the force applied to the object times the distance the object is moved while the force is applied or, more succinctly, work = (force)(distance). We will use the International System of Units: Force is measured in Newtons, N, and distance is measured in meters, m. Weight is the gravitational force applied to the object and cable, therefore, weight = force. Cable (3 N per m) 40 - X Object (1500 N) Ax How much work is done lifting a 1500 N object from the ground to the top of a 40 m building if the cable weighs 3 N per m? Here the 1500 N object is constant, but as you retract the cable, the overall weight (force) decreases. The diagram to the right can help you derive the integral to calculate the work done from the ground to the top of the buildingX1=40X2=0

Answer:

electromagnetic waves only

Explanation:

I just took the test, Hope it helps!

Answer:

A: Electromagnetic waves only

Explanation:

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Answer:

7.23407 [tex]\frac{m}{s^2}[/tex]

Explanation:

(I will not include units in calculations)

I'm assuming FBD's are already drawn, so I will work from there.

Let the 2.2kg block equal [tex]m_2[/tex], and the 20kg block equal [tex]m_1[/tex].

Summation equation for [tex]m_2[/tex]: [tex]\sum F_x=F_t_2-(F_f+F_g_x)=m_2a[/tex], [tex]\sum F_y=F_n-F_g_y=0[/tex]

Summation equation for [tex]m_1[/tex]: [tex]\sum F_y=F_g-F_t_1=m_1a[/tex]

Torque Summation Equation: [tex]\sum\tau=F_t_1*r-F_t_2*r=I\alpha[/tex]

Do some plugging in with the values given: [tex]\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha[/tex]

Replace [tex]\alpha[/tex] with [tex]\frac{a}{r}[/tex], and cancel out the r's.

[tex]\sum\tau=F_t_1-F_t_2=.5Ma[/tex]

This step is important: Rearrange the force summation equation to solve for each tension force.

[tex]F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a[/tex]

Perform Substitution: [tex]\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma[/tex]

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that [tex]F_f=[/tex]μ[tex]F_n[/tex]

And based on our earlier summation equation: [tex]F_n=F_g_y[/tex]

First, break [tex]F_g[/tex] into x and y components. [tex]F_g_y=F_g\cos(\theta)[/tex], [tex]F_g_x=F_g\sin(\theta)[/tex]

Perform substitution with this and the fact that [tex]F_g=mg[/tex].

[tex]\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma[/tex]

Solving for a, plugging in numbers yields an answer of 7.23407 [tex]\frac{m}{s^2}[/tex]

Answer:

7.23407

Explanation:

easy

Answer:  lose weight at high altitudes.

Explanation:

Answer:

Just a week at high altitudes can cause sustained weight loss, suggesting that a mountain retreat could be a viable strategy for slimming down. Overweight, sedentary people who spent a week at an elevation of 8,700 feet lost weight while eating as much as they wanted and doing no exercise

The frequency of the waves will be 18 Hz. Then the correct option is B.

The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. One event occurs per second when measuring frequency in hertz.

Frequency is the number that represents the number of oscillations or intervals each second. The hertz is the SI unit for frequency (Hz). One cycle per second equals one hertz.

The number of finished waves produced each second is considered the wave frequency.

Six wave crests pass a fisherman sitting at the end of a pier in the span of three seconds.

The frequency of the waves is given as,

f = 6 x 3

f = 18 Hz

The waves will have a frequency of 18 Hz. Then, choice B is the best one.

More about the frequency link is given below.

https://brainly.com/question/29739263

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If everything else remains constant, then the received signal power is simply an inverse-square function of its distance from the transmitter.

That is, down 6dB when distance is doubled etc.

If distance increases from 16 to 20, then received power decreases by the factor of (16/20)^2.

That's (0.8)^2 = 0.64

New receive power is 5x0.64 = 3.2 nW

=. ===== ==========

Here's how I do it at my job:

Initial RSL = 5nW ~ - 53 dBm

loss = 20log(20/16)= 20log(1.25)~1.94dB

New RSL = - 54.94 dBm or ~ 3.2 nW .

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:

The Kinetic energy of Sphere is higher than the cylinder.

( KS > KC )

Explanation:

Given - A sphere with the same mass and radius as the original cylinder, but a smaller rotational inertia, is released from rest from the top of the ramp. KS and KC are the sphere's and cylinder's total kinetic energy at the bottom of the ramp, respectively.

To find - How do KS and KC compare, and why ?

Proof -

We know that,

The total energy of an object =  Potential energy +  linear kinetic energy + rotational kinetic energy.

⇒E = mgh + [tex]\frac{1}{2} mv^{2}[/tex] + [tex]\frac{1}{2} l\omega^{2}[/tex]  

Now,

Mass of sphere = m

Radius of sphere = r

So,

The moment of inertia of a uniform solid sphere = [tex]\frac{2}{5} mr^{2}[/tex]

Also,

Mass of cylinder = m

Radius of cylinder = r

So,

The moment of inertia of a uniform solid cylinder = [tex]\frac{1}{2} mr^{2}[/tex]

Now,

Total energy for the sphere , Es = mgh + [tex]\frac{7}{10} mv^{2}[/tex]

Total energy for the cylinder, Ec =  mgh + [tex]\frac{3}{4} mv^{2}[/tex]

As they always have the same total energy,

So, for height h of the sphere's velocity has to be higher.

Therefore,

The Kinetic energy of Sphere is higher than the cylinder.

Answer:

KS < KC

Explanation:

Answer:

1

Explanation:

Answer:

636.6 W/m²

Explanation:

From the given information:

The area of the circular spot can be calculated as:

A = πr²

A = π(0.5 × 10⁻³ m)²

A = 7.85 × 10⁻⁷ m²

The intensity can be determined by using the formula:

[tex]I = \dfrac{P}{A}[/tex]

[tex]I = \Big ( \dfrac{0.500 \ mW}{\pi (0.5 \times 10^{-3} \ m )^2} \Big) \Big( \dfrac{10^{-3} \ W}{1\ mW} \Big)[/tex]

[tex]I = \dfrac{0.500 \times 10^{-3} }{\pi(0.25 \times 10^{-6} )}[/tex]

I = 636.6 W/m²

Answer:

w = 6.325 rad/s

Explanation:

Given the following data;

Mass = 2kg

Radius, r = 10m

Centripetal force = 800N

To find the angular velocity of the car;

First of all, we would use the following formula to solve for the speed of the car;

Centripetal force = mv²/r

800 = 2*v²/10

Cross-multiplying, we have;

8000 = 2v²

4000 = v²

Taking the square root of both sides, we have;

v = √4000

v = 63.25 m/s

Next, we find the angular velocity, w using the formula below;

w = v/r

w = 63.25/10

w = 6.325 rad/s

Answer:

ωf = 113.95 rev/s

t = 1.26 s

Explanation:

We can use the third equation of motion to find out the final spinning speed of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_i^2\\[/tex]

where,

α = angular acceleration = 65 rev/s²

θ = No. of revolutions completed = 92 rev

ωf = final angular speed = ?

ωi = initial angular speed = 32 rev/s

Therefore,

[tex](2)(65\ rev/s^2)(92\ rev) = \omega_f^2 - (32\ rev/s)^2\\\omega_f^2 = 11960\ rev^2/s^2 + 1024\ rev^2/s^2\\\omega_f = \sqrt{12984\ rev^2/s^2}[/tex]

ωf = 113.95 rev/s

Now, for the time we can use the first equation of motion:

[tex]\omega_f = \omega_i +\alpha t\\113.95\ rev/s - 32\ rev/s = (65\ rev/s^2)t\\t = \frac{81.95\ rev/s}{65\ rev/s^2}\\\\[/tex]

t = 1.26 s

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M[tex]_S[/tex] = 1.99 × 10³⁰ kg

Mass of the neutron star

M[tex]_N[/tex] = 2( M[tex]_S[/tex] )

M[tex]_N[/tex] = 2( 1.99 × 10³⁰ kg )

M[tex]_N[/tex] = ( 3.98 × 10³⁰ kg )

Radius of neutron star R[tex]_N[/tex] = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω[tex]_N[/tex].

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM[tex]_N[/tex] = / R[tex]_N[/tex]² = mR[tex]_N[/tex]ω[tex]_N[/tex]²

ω[tex]_N[/tex]² = GM[tex]_N[/tex] = / R[tex]_N[/tex]³

ω[tex]_N[/tex] = √(GM[tex]_N[/tex] = / R[tex]_N[/tex]³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω[tex]_N[/tex] = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω[tex]_N[/tex] = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω[tex]_N[/tex] = √ 120831133.3636777

ω[tex]_N[/tex] = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

Answer:

pizza

Explanation:

Answer:

sea food???

Explanation:

Answer:

A·B = 4

Explanation:

Given that,

Vector A = 2i+3j-3k

Vector B = -i+5j+3k

We need to find the value of A·B.

We know that,

i·i=j·j=k·k = 1 and i·j=j·k=k·i=0

So,

[tex]A\cdot B=(2i+3j-3k)\cdot (-i+5j+3k)\\\\=-2+5(3)+(-3)(3)\\\\=-3+15-9\\\\=4[/tex]

So, the value of A·B is equal to 4.

Answer:

heat and gas sometimes so true

Answer: 0.83 m

Explanation:

Given

mass of the block is m=3 kg

spring constant k=98 N/m

The Speed at the time of collision is v=1.5 m/s

Here, the kinetic energy of the block is converted into Elastic potential energy

[tex]\Rightarrow \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2 \\\\\Rightarrow 3\times 1.5^2=98\times x^2\\\\\Rightarrow x^2=0.6887\\\\\Rightarrow x=0.829\approx 0.83\ m[/tex]

Answer:

(a) m³

Explanation:

Joule is the unit of work and Pascal is unit of pressure.

J/Pa = work/pressure = Nm/Nm⁻² = m³

Thus, The unit J/Pa is equivalent to m³

-TheUnknownScientist

Answer:

1st and 2nd

Explanation:

Answer:

Diagram (3).

Explanation:

N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).

The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).

The law clearly states a Force of attraction; the two objects come towards each other.

Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.

Read more:

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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:

The correct answer is - option E. e. F1 > F2 and W1 = W2

Explanation:

Case 1 - F1 = mg, m= mass and g = gravitational force

Work done W1 = F1h= mgh (h - height of bed of truck)

Case 2 - F2 = mgsinθ

work done W2 = mgsinθ×L (L=length of ramp ans sin θ angle of ramp)

sinθ = h/l, then W2 = mg(h/L)L

W2 = mgh

By comparing both,

F2 = mgsinθ and F1 = mg therefore,

F2>F1

now, W2 = mgh, and W1 = mgh

therefore, W2 = W1.

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:

the maximum current is 500 A

Explanation:

Given the data in the question;

the B field magnitude on the surface of the wire is;

B = μ₀i / 2πr

we are to determine the maximum current so we rearrange to find i

B2πr = μ₀i

i = B2πr / μ₀

given that;

diameter d = 2 mm = 0.002 m

radius = 0.002 / 2 = 0.001 m

B = 0.100 T

we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A

so we substitute

i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷

i = 500 A

Therefore, the maximum current is 500 A

Answer:

a = -3984.6 m/s²

F = 577.76 N

Explanation:

The acceleration of the ball can be calculated by using the third equation of motion:

[tex]2as = v_f^2 - v_i^2\\[/tex]

where,

a = acceleration of ball = ?

s = distance covered = recoil distance = 0.135 m

vf = final speed = 0 m/s

vi = initial speed = 38.2 m/s

Therefore,

[tex]2(0.135\ m)a = (0\ m/s)^2-(38.2\ m/s)^2\\[/tex]

a = -3984.6 m/s²

here negative sign shows deceleration.

Now, for the force applied by Brian Lara will be equal in magnitude but opposite in direction of the force required to stop the ball:

[tex]F = -ma\\F = -(0.145\ kg)(-3984.6\ m/s^2)\\[/tex]

F = 577.76 N

Answer:

it's A subduction, deep water trench

Answer:

none no work cuz no motion

Explanation:

GOOD LUCK