A constant horizontal force of 30.0 N is exerted by a string attached to a 5.0 kg block being pulled across a tabletop. The block also experiences a frictional force of 5.0 N due to contact with the table. What is the horizontal acceleration of the block?​

Answer:

●Bx=Bcos40

Bx=10 × 0.76

Bx=7.6

●By=Bsin40

By=10×0.64

By=6.4

Hope it will help you.

[tex] \large★·.·´¯`·.·★ {Answer}★·.·´¯`·.·★[/tex]

As we know that Kinetic Energy is the Energy that is possessed by a moving object. and if the object is at rest then it doesn't have velocity therefore there is no kinetic Energy.

In the numerical terms we can express it as : -

[tex]꧁  \:  \large \frak{Eternal \:  Being } \: ꧂[/tex]

W = 25 J

Explanation:

Work done on an object is defined as

[tex]W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}[/tex]

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

Answer:

Explanation:

F = GMm/d²

d = √(GMm/F)

d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)

d = 1.29398e8 = 1.29 x 10^8 m  center to center

Answer:

1.29 x 10^8 m  apart

Explanation:

Works in Acellus!

Answer:

Explanation:

Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.

Let m be the mass of the car and ground level is the origin.

on the 5.5 m hill, total energy is

E = PE + KE

E = mgh + ½mv²  

E = m(9.8)(5.5) + ½m(9.3)² = 97m J

a) The maximum height will occur when the total energy is all potential energy.

E = mgh

h = E/mg

h = 97m/m(9.8) = 9.9 m  

As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.

b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.

c) KE = E - PE

KE = 97m - m(9.8)(9.5 - 1.0)

KE = 97m = 83.3m

KE = 13.7m = ½mv²

v² = √(2(13.7)

v = 5.2345...

v = 5.2 m/s

Hi there!

We can begin by finding the period of the pendulum.

[tex]T = \text{ # of complete swings / seconds} = 12 / 8 = \boxed{\text{1.5 sec}}[/tex]

The frequency is simply the reciprocal of the period, so:

[tex]f = \frac{1}{T} = \frac{1}{1.5} = \frac{2}{3}Hz \text{ or } \boxed{0.67 Hz}[/tex]

Answer:

Pick c is the right one

Explanation:

5.0m/s

Answer:

TRUE

FALSE

TRUE

TRUE

TRUE

TRUE

Explanation:

1. Most EM waves has the same speed. True

2. EM waves travel at the speed of 4x108 m/s. False

3. Electromagnetic waves are transverse waves consisting of changing electric fields and changing magnetic fields. True

4. Electromagnetic waves transfer energy through a vacuum. True

5. A wave is a disturbance that transfers energy. True

Answer:

D = 25g/cm³

Explanation:

1ml = 1cm³

D = m/V

D = 500g/20cm³

D = 25g/cm³

Answer:

currents which circulate in conductors like swirling eddies in a stream in electromagnetic series....

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

Answer:

10.83 Amperes

Explanation:

if   A ⇒ current

W = VA

1300 = 120 x A

1300 / 120 = A

10.83 = A

Answer:

0

Explanation:

0 is the answer

Answer:

Explanation:

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

A 100 N create is being pulled at a constant velocity by a rope a 30 degrees to the horizontal as depicted in the diagram given in question the force of friction Ff = 100μcos30 / (cos30 + ½μ).

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

the force of friction Ff, is 100μcos30 / (cos30 + ½μ).

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ5

Answer:

it's ahfdfhhh hhgfdjjjjuyggffdddcff

Answer:

Summertime

Explanation:

the sun never sets south of the Antarctic circle in the summertime.

Answer:

everything will float up and go up to space and die

Explanation:

gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.

Answer:

Explanation:

If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change

Explanation:

The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as

[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]

where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.

Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write

[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]

[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]

where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is

[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]

[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]

Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,

[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]

or

[tex]4\pi^2r = gT^2[/tex]

Solving for T, we get

[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]

We can further simplify the above expression into

[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]

Plugging in the values for r and g, we get

[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]

Answer:

The Aluminum

Explanation:

With a larger specific heat, water requires more heat to raise its temperature by a temperature degree.

In this system, with equal masses of water and aluminum, the heat moving from the aluminum lowering its temperature by one degree is not sufficient to raise the water temperature by one degree.

Answer:

a vacuum tube is NOT a solid state component