Help me answer this please!

Answer:

b

Explanation:

Answer:

444.99

Explanation:

Explanation:

C=A-B

=(4.00i+7.00j)-(5.00i-2.00j)

= -1.00i+9.00j

A.B=AxBx+AyBy

=(4 x 5)+(7 x -2)

=20-14

=6

Answer:

The answer is C. A force of 130 N moves it 22 m

The least amount of work done on a crate is when A force of 130 N moves it 22 m, the correct option is C.

The total amount of energy transferred when a force is applied to move an object through some distance

The work done is the multiplication of applied force with the displacement.

Work Done = Force * Displacement

The work done for option A is

                = 120*25= 3000 Nm

The work done for option B is

                 =115*26 = 2990 Nm

The work done for option C is

                 =130*22= 2860 Nm

The work don for option D is

                 = 125*27= 3375 Nm

Hence, the least amount of work is done on the crate when a force of 130 N moves it 22 m.

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Answer: C

Explanation: Amplitude controls loudness, and frequency controls pitch. The more frequent the higher pitch.

The true statement regarding the two sounds is wave A has a lower pitch.

The pitch of a sound is sensation of the sound frequency.

A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to low frequency.

Sound of the the same amplitude, have the same loudness.

Thus, the true statement regarding the two sounds is wave A has a lower pitch.

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Answer:

Both cars convert more kinetic energy to thermal energy.

Explanation:

Given that,

Two cars come to a stop from the same initial speed, one braking gently and the other braking hard.

We know that,

The conservation of energy :  

The energy of the system is always constant.

The energy can be change one form to another form.

So, we can say that when both cars come to a stop from the same speed, one braking softly and the other braking strongly then both cars convert more kinetic energy to thermal energy.

Hence, Both cars convert more kinetic energy to thermal energy.

Based on the law of consevation of energy, the car that brakes hard will convert more kinetic energy to thermal energy.

According to the law of conservation of energy, the total energy is conserved in any closed system.

Since the cars have the same mass and had the same initial speed, their kinetic energies are the same.

However, the car braking hard comes to rest in a shorter time and in a shorter distance.

Using the formula for calculating power:

The car that braked hard will have greater power due to shorter time of action.

This power is seen as greater thermal energy, whereas some of the energy of the other car will still be used in travelling some distance before stopping.

Therefore, the car that brakes hard will convert more kinetic energy to thermal energy.

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Answer:

Explanation:

The work done by an object can be found by using the formula

From the question

force = 290 N

distance = 5 m

We have

workdone = 290 × 5

We have the final answer as

Hope this helps you

Answer:

3.2

Explanation:

I hope that this helps! Have a good day!!

Answer:

The boat has more buoyancy

Explanation:

Answer:

Geocentric model of the solar system helped to explain retrograde motions of planets. Explanation: Geocentric model of planets was proposed by Ptolemy. It stated that all sun, planets and stars revolve round the earth in circular orbits.

Answer:

retrograde motion

Explanation:

i reverse searched the image

(a) Differentiate the position vector to get the velocity vector:

r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k

v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j

(b) The velocity at t = 2.00 s is

v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j

(c) Compute the electron's position at t = 2.00 s:

r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k

The electron's distance from the origin at t = 2.00 is the magnitude of this vector:

||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that

tan(θ) = (-16.0 m/s) / (3.00 m/s)   ==>   θ ≈ -79.4º

or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

Answer:

Humans be considered carbon sinks. Not only do humans

have a lot of carbon in them, they also use a lot of carbon.

hope this helps

Answer:

The larger object will have smaller  acceleration that the less massive object.

Explanation:

Generally force is mathematically represented as

      [tex]F = ma[/tex]

=>  [tex]m = \frac{F}{a }[/tex]

at constant  force  we have

     [tex]m \ \alpha \ \frac{1}{a}[/tex]

So if  m is  increasing a will be decreasing which means the object with the larger mass will have less acceleration

Answer:

M V = m v        conservation of momentum (Caps-cannon  Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s    recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2  = 40 kg m^2/s^2 = 40 J

Answer:

The force of forearm is 239.9 N

The force of elbow is 215.89 N

Explanation:

Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance [tex]d_{b}[/tex] 2.50 cm from the elbow, how much force [tex]F_{b}[/tex] must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force [tex]F_{l}[/tex] must the elbow exert,

Given that,

Mass of forearm = 1.50 kg

Distance of forearm = 2.50 cm

Mass of ball = 950 g

Distance of ball = 36.0 cm

We need to calculate the force of forearm

Using balancing torque about elbow

[tex]F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}[/tex]

Put the value into the formula

[tex]F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36[/tex]

[tex]F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}[/tex]

[tex]F_{b}=239.9\ N[/tex]

We need to calculate the force of elbow

Using balancing force

[tex]F_{b}=F_{l}+w_{f}+w_{b}[/tex]

[tex]F_{l}=F_{b}-w_{f}-w_{b}[/tex]

Put the value into the formula

[tex]F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)[/tex]

[tex]F_{l}=215.89\ N[/tex]

Hence, The force of forearm is 239.9 N

The force of elbow is 215.89 N

Explanation:

classical physics and modern physics.

Physics is the branch of science that deals with the interaction of energy and matter. Modern and Classical Physics are the two major branches of physics.

Physics is the discipline of science that investigates the structure of matter and how the universe's fundamental constituents interact. It investigates objects ranging from the very small to the entire universe using quantum mechanics and general relativity.

One of the most crucial physics topics is kinematics. Kinematics is the study of an object 's direction of travel. Kinematics is only concerned with the motion of an object.

There is a vertical and a horizontal component to two-dimensional projectile motion, such as that of a football or other thrown object.

Throwing a rock or kicking a ball produces a projectile pattern of motion that includes both a vertical and a horizontal component.

Thus, these can be the topics that require the knowledge of physics.

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Complete question is;

Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at

(x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)

Compute the length of the bungee cord at t = 7 seconds. (Round your answer to three decimal places.)

Answer:

Length of bungee cord = 85.734 ft

Explanation:

We are told that Adrian moves 9ft/sec. Thus, at 5.5 seconds, distance he moved is; 9 ft/sec × 5.5sec = 49.5 ft in the negative x (-x) direction. Therefore, the coordinate is (-49.5, 0).

Now, Allyson has moved 10ft/sec. Thus, at 7 seconds, distance he moved would be; 10 ft/sec x 7sec = 70 feet in the positive (+y) direction. Therefore, the coordinate is (0, 70).

Now, since they started from the origin, it means (0, 0) is a coordinate. Thus, we now have 3 coordinates which are; (0, 0), (0, 70) & (-49.5,0). These 3 coordinates would therefore combine to form a right triangle.

The hypotenuse is the distance between Allyson and Adrian.

Thus, from pythagoras theorem, we can find the distance between them which is same as the length of the cord.

Thus;

(-49.5)² + 70² = D².

D² = 2450.25 + 4900

D = √7350.25

D = 85.734 ft

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

[tex]v(t) =\frac{d(f(t))}{dt}[/tex]

[tex]v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2[/tex]

Hence the velocity of the particle at time t is [tex]v(t) = 0.04t^3 - 0.09t^2[/tex]

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

[tex]v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05[/tex]

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second

Answer:

The options are not shown, so i will answer in a general way:

We can separate the problem in two cases.

The movement in the x-axis, that is unperturbed, so the particle will keep moving in the +x direction with the same speed.

The movement in the y-axis, as a constant force is applied, by the second Newton's law:

F = m*a

a = F/m = constant

v = a*t + v0

p = (a/2)*t^2 + v0*t + p0

Where v0 = initial velocity in the y-axis, p0 = initial position (in this case the origin) and t is the time. (In this case, these values really do not matter, i only want to show the general equations of movement in the y-axis)

We will also have a constant acceleration, which means that the velocity in this axis will be linear, and the position quadratic, then after the particle reaches the origin, it will start moving up and to the right, and the upward movement will increase in magnitude as time passes, a graph in the x-y plane would be something like:

The amount of kinetic energy in an object can be increased by two ways:

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.

The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest.

Mathematically kinetic energy can be expressed as:

Kinetic energy = 1/2 × mass × speed²

Hence, the amount of kinetic energy in an object can be increased by two ways:

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Answer:

The momentum of the boat A is 10 kg.m/s.

Explanation:

Given;

velocity of the two boats, V = 1 m/s

mass of boat A, mₐ = 10 kg

mass of boat B, mb = 5 kg

Linear momentum is given by the product of mass and linear velocity.

The momentum of the boat A is given by;

Pₐ = mₐv

Pₐ = 10 x 1

Pₐ = 10 kg.m/s

Therefore, the momentum of the boat A is 10 kg.m/s.

Complete Question

Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:

τs = -κ θ.

Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.

Required:

a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.

b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device? Assume the full scale deflection is 60° from the spring's equilibrium position

Answer:

a

[tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

b

[tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

Explanation:

From the question we are told that

   The number of turns is  N  =  50  

   The radius is  r =  1.55 cm  =  0.0155 m

   The  magnetic field is  B  =  0.35 T

   The induced current is  [tex]I =  45 \mu A =  45 *10^{-6} \  A[/tex]

Generally the area of  loop is mathematically represented as

      [tex]A =  \pi r^2[/tex]

=>   [tex]A =3.142 *  0.0155^2[/tex]

=>   [tex]A =0.000755\ m^2[/tex]

Generally the maximum torque is mathematically represented as

     [tex]\tau_{m} =  N  *  B  * I  *  A[/tex]

=>  [tex]\tau_{m} =  50  *  0.35  * 45 *10^{-6} *  0.000755[/tex]

=>  [tex]\tau_{m} =  5.95 *10^{-7} \  N \cdot m[/tex]

Generally the  torque 60° from the spring's equilibrium position is mathematically represented as

      [tex]\tau  =  N  *  B  *  I  *  A  *  sin (60)[/tex]

=>    [tex]\tau  =  50  *  0.35  *  45 *10^{-6} *  0.000755  *  sin (60)[/tex]

=>   [tex]\tau  = 2.973 *10^{-7} \  N  \cdot m [/tex]

Generally the toque constant of the spring is mathematically represented as

     [tex]\beta =  \frac{\tau}{60}[/tex]

=>   [tex]\beta =  \frac{\tau}{\frac{\pi}{3}}[/tex]

=>   [tex]\beta =  \frac{2.973 *10^{-7}}{\frac{\pi}{3}}[/tex]

=>    [tex]\beta = 2.83 *10^{-7} \  N  \cdot m / rad [/tex]

Answer:

μ_k = 0.851

Explanation:

We are given;

Mass of book; m_book = 4.7 kg

Horizontal force; F_horiz = 42 N

Distance; d = 0.85 m

Speed; v = 1 m/s

First of all let's find the acceleration using Newton's equation of motion;

v² = u² + 2ad

u is initial velocity and it's 0 m/s in this case.

Thus;

1² = (2 × 0.85)a

1 = 1.7a

a = 1/1.7

a = 0.5882 m/s²

Now, resolving forces along the vertical direction, we have;

W - N = 0

Thus,W = N

Where W is weight = mg and N is normal force

Thus; N = mg = 4.7 × 9.81 = 46.107 N

Now, resolving forces along the horizontal direction, we have;

F_horiz - ((μ_k)N) = ma

Where μ_k is coefficient of kinetic friction.

Thus;

42 - 46.107(μ_k) = 4.7 × 0.5882

42 - 46.107(μ_k) = 2.76454

μ_k = (42 - 2.76454)/46.107

μ_k = 0.851

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equation is   [tex]c = \frac{1}{\sqrt{ \mu_o *  \epsilon_o} }[/tex]

The value of  c is [tex]c = 2.998 *10^{8} \  m/s  [/tex]

Explanation:

From the question we are told that

   Generally the equation that  relates the speed of light to other fundamental electromagnetic constants is

     [tex]c = \frac{1}{\sqrt{ \mu_o *  \epsilon_o} }[/tex]

Here  c is the speed of light

[tex]\mu_o[/tex] is the permeability of free space with value

    [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]

and  [tex]\epsilon_o[/tex] is the permittivity of free space  with value  

      [tex]\epsilon_o  =  8.85*10^{-12} \ C/V \cdot m[/tex]

So

      [tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7}  *  8.85*10^{-12}} }[/tex]

=>   [tex]c = 2.998 *10^{8} \  m/s  [/tex]

Answer:

The angle is 3.95 rad.

Explanation:

The angle can be calculated as follows:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]

Where:

[tex]\omega_{f}[/tex]: is the final angular speed

ω₀: is the initial angular speed = 0 (it starts from rest)

α: is the angular acceleration

θ: is the angle=?

The centripetal acceleration is:

[tex]a_{c} = \omega_{f}^{2}*r[/tex]

And the tangential acceleration is:

[tex] a_{T} = \alpha*r [/tex]

Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:

[tex]a_{c} = 7.9a_{T}[/tex]

[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]

Now, the angle is:

[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]

[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]

Therefore, the angle is 3.95 rad.

I hope it helps you!          

The angular distance traveled by the electric drill is 3.95 radians.

The given parameters;

The angular distance traveled by the electric drill is calculated as follows;

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

The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;

[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]

Substitute the value of angular acceleration into the first equation;

[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]

Thus, the angular distance traveled by the electric drill is 3.95 radians.

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