A custodian pulls a vacuum 13.5 m
with a 33.9 N force at a 55.0°
angle, against a 14.2 N friction
force. Find the total work done on
the vacuum.
(Unit = J)

Answer:

27 ms^-1

Explanation:

by using v= u + at

u = 0 ( because the object id starting from rest)

v= 0 + 1.5 x 18

v = 27 ms^-1

Answer:

You must label the horizontal axis with the names of the airlines and the vertical axis with the number of flights. The title must clearly state what data the bar chart is showing. With larger numbers, your scale may not go up by one.

Answer:

You must label the horizontal axis with the names of the airlines and the vertical axis with the number of flights. The title must clearly state what data the bar chart is showing. With larger numbers, your scale may not go up by one.

Explanation:

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Answer:

stop

Explanation:

object stop

Answer:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement. Two or more opposite forces are balanced forces if their effects cancel each other and they do not cause a change in an object's motion.

Answer:

one sec let me think

Explanation:

(a)The average velocity of the ball over the following time intervals will be  [3,4] ft/sec.

(b)The instantaneous velocity at time t = 3 will be32 ft/sec.

(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)The ball will hit the ground at 13.4 sec.

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity. it is a time-based component.

The given data in the question will be ,

u is the initial velocity by which ball thrown=128 ft/sec.

V₃ is the instantaneous velocity at time t=3 sec.

V₆ is the instantaneous velocity at time t=6 sec.

t is the time when ball hits the ground=?

(a) Given equation for the displacement

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

Time when velocity is zero will be

[tex]\rm{ t=\frac{128}{32}[/tex]

[tex]\rm{ t=4 sec[/tex]

If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .

Hence the average velocity obtained from the problem will be  [3,4] ft/sec

(b)  

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=3 sec)

v(3) = 128-32×3

v(3) =32 m/sec.

Hence the instantaneous velocity at time t = 3 will be32 ft/sec.

(c)

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=6 sec)

v(6) = 128-32×6

v(6) = -64 m/sec.

Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)

According to Newtons third equation of motion we got

v=u+gt

If the body returens from a certain height at max height its velocity must be zero; ( u=0)

[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]

Hence the ball will hit the ground at 13.4 sec.

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Answer:

20mm per second

Explanation:

lol

Answer:

2.2 ma

Explanation:

Given :

Length of the rope = L

Mass of the rope = m

Mass of the object pulled by the rope = M

M = 1.5 m

So, L [tex]$\rightarrow$[/tex] m

For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]

∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.3 m

And for remaining 0.7 L =  [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.7 m

By Newtons law of motion,

F - T = ( 0.3 m) a .........(1)

T = ( M + 0.7 m) a

T = ( 1.5 m + 0.7 m) a

T = ( 2.2 m ) a  ..............(2)

So from equation (1) and (2), we have

Tension on the rope

T = 2.2 ma

magnetic domain is a region within a magnetic material in which the magnetization is in a uniform direction. This means that the individual magnetic moments of the atoms are aligned with one another and they point in the same direction.

Answer:

[tex]31.035^{\circ}[/tex]

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Displacement in x direction is given by

[tex]x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}[/tex]

Displacement in y direction is given by

[tex]y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}[/tex]

[tex]\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}[/tex]

The angle at which the ball was thrown is [tex]31.035^{\circ}[/tex].

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

Answer: 0.33 m/s^2

Explanation:

The acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²

Given the data in the question;

To determine the acceleration of the ship, We the equation from Newton's Second Law of Motion:

[tex]F = m\ *\ a[/tex]

Where F is the force, m is the mass and a is the acceleration

Lets make acceleration ''a'', the subject of the formula

[tex]a = \frac{F}{m}[/tex]

Now, we substitute our given values into the equation

[tex]a = \frac{300000N}{900000kg}[/tex]

We know that, A newton is defined as [tex]1 kg.m/s^2[/tex]

[tex]a = \frac{300000 kg.m/s^2}{900000kg} \\\\a = 0.33m/s^2[/tex]

Therefore, the acceleration of Odysseus' ship as it moves across the water from Troy to Ithaca is 0.33m/s²

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Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg  We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

[tex]F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N[/tex]

So, the net force is 2985.125 N.

Answer:

 Δt = 5.85 s

Explanation:

For this exercise let's use Faraday's Law

           emf = [tex]- \frac{d \phi}{dt}[/tex] -  d fi / dt

           [tex]\phi[/tex] = B. A

           \phi = B A cos θ

The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.

suppose a linear change of the magnetic field

            emf = - A [tex]\frac{B_f - B_o}{ \Delta t}[/tex]

             Dt = - A  [tex]\frac{B_f - B_o}{emf}[/tex]

the final field before a fault is zero

let's calculate

            Δt = - 0.046 (0- 1.4) / 0.011

            Δt = 5.85 s

Answer: Option b.

Explanation:

We know:

Wavelength = 632.8 nm

Fluence = 4.5*10^20 photon/s

The energy of a single photon of wavelength λ is:

E = (h*c)/λ

where:

h = 6.6*10^(-34) J*s

c = 3*10^8 m/s

And we should rewrite the wave length in meters, so:

λ = 632.8 nm = 632.8*10^(-9) m

replacing these in the energy equation, we get:

E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J

So each one of the  4.5x10^20 photon that flow each second have this energy, then the power is:

P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s

and 1 W = 140.85 J/S

Then the power is:

P = 140.85 W

Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.

Answer:

[tex]2.4375\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of cannon ball = 39 kg

[tex]m_2[/tex] = Mass of performer = 65 kg

[tex]u_1[/tex] = Initial horizontal component of cannon ball's velocity = 6.5 m/s

[tex]u_2[/tex] = Initial horizontal component of performer's velocity = 0

v = Velocity of combined mass

As the momentum of the system is conserved we have

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}[/tex]

The speed of the performer immediately after catching the cannon ball is [tex]2.4375\ \text{m/s}[/tex].

Answer:

true

Explanation:

because I tride it

Answer:

The reckage after collision is motionless (E)

Explanation:

The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.

The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.

The calculation is as follows:

m₁v₁ + m₂v₂

where:

m₁ = m₂ = 1000kg

v₁ = 20 m/s

v₂ = -20 m/s ( in the opposite vector direction)

∴ resultant momentum = (1000 × 20) + (1000 × -20)

= 20000 - 20000 = 0

∴ The reckage after collision is motionless

Answer:

The wreckage after collision is moving at the speed 18 m/s to the south.

Explanation:

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ([tex]r[/tex]), measured in kilometers, can be represented by a right triangle:

[tex]r = \sqrt{x^{2}+y^{2}}[/tex] (1)

Where:

[tex]x[/tex] - Horizontal distance between the rocket and the observer, measured in kilometers.

[tex]y[/tex] - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ([tex]\theta[/tex]), measured in sexagesimal degrees, is defined by the following trigonometric relation:

[tex]\tan \theta = \frac{y}{x}[/tex] (2)

If we know that [tex]x = 5\,km[/tex], then the expression is:

[tex]\tan \theta = \frac{y}{5}[/tex]

And the rate of change of this angle is determined by derivatives:

[tex]\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y[/tex]

[tex]\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}[/tex]

[tex]\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}[/tex]

[tex]\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}[/tex]

Where:

[tex]\dot \theta[/tex] - Rate of change of the angle of elevation, measured in sexagesimal degrees.

[tex]\dot y[/tex] - Vertical speed of the rocket, measured in kilometers per hour.

If we know that [tex]y = 4\,km[/tex] and [tex]\dot y = 400\,\frac{km}{h}[/tex], then the rate of change of the angle of elevation is:

[tex]\dot \theta = 48.780\,\frac{\circ}{s}[/tex]

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Answer:

B = An 8 kg book at a height of 3 m

Answer:

B is the Answer

Follow me please

Mark brainliest

Answer:

The answer is A

Explanation:

This is because f = 1/T,

or frequency equals 1 over period, T

Since the highest T would be the answer with the greatest length and mass does not matter the answer would be c IF they were asking for the highest period.

F is found by 1 over T so the greatest F would be the smallest number, Hence answer A

Please Mark me brainiest

Answer:

2500 m

Explanation:

Given that,

Each lap is 500 m

Gareth takes 5 laps.

We need to find distance traveled by Gareth. The distance covered by him is given by :

d = 5×500

d = 2500 m

Hence, he will travel 2500 m.

Answer:

The sun and the stars

Explanation:

I hope this helps!

Answer:

Northstar & North Pole

Explanation:

yyggggggggggggggg

Answer:

The value 0.51

Explanation:

Answer:

Bowling ball

Explanation:

Answer: Option b, the final velocity is half of the initial velocity.

Explanation:

Here we will use the conservation of the total momentum of a system.

This means that the total momentum at the beginning must be the same as the final momentum.

Where momentum is:

P = M*v

Initially, we have two cars, both with the same mass M, and only one of them has a velocity v.

Then the initial momentum is:

P = M*v + M*0 = M*v

After the collision, the two cars move together. Then the total mass that is moving is equal to the sum of the masses of the cars, this is 2*M

and we can suppose that the two cars move at a final velocity v'

Then the final momentum is:

P' = (2*M)*v'

Now we use the conservation of momentum, then:

P = P'

M*v = (2*M)*v'

Now we need to solve this for v'

(M*v)/(2*M) = v'

v/2 = v'

This means that the final velocity is half of the initial velocity.

Then the correct option is option b.