pls solve this:

if u solve this, I will surely mark u as brilliant

Answer:

the work done by the motor is 1,044 J.

Explanation:

Given;

the output power of the motor, P = 0.7 hp

duration of the work, t = 2 s

The relationship between horse-power and watt is given as;

1 hp = 745.7 W

0.7 hp = ?

0.7 hp = 522 W = 522 J/s

The work done by the motor is calculated as;

W = Power x time

W = 522 J/s  x  2 s

W = 1,044 J

Therefore, the work done by the motor is 1,044 J.

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

Given the following data;

Radius, r = 2.6 km

Time = 360 seconds

Conversion:

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]

Where;

Substituting into the formula, we have;

[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]

[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]

Where;

Substituting into the formula, we have;

[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]

[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]

Centripetal acceleration = 0.79 m/s²

Answer:

(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J

(b) the efficiency of the athlete is 10.94 %

Explanation:

Given;

work done by the athlete (system), W = 6.53 x 10⁴ J

the heat given off by the athlete (system), Q = 5.97 x 10⁵ J

The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.

                            Q← ⊕ →W

The work, W, points away from the system since the system does the work

The heat, Q, points away from the system since heat is given off

Apply first law of thermodynamic;

ΔU = Q + W

where;

q is the heat flowing into or out of the system

(+q     if the heat is flowing into the system

(-q      if the heat is leaving the system

w is the work done by or on the system

(+w     if the work is done on the system by the surrounding

(-w     if the work is done by the system to the surrounding

Thus, from the above explanation, the change in internal energy of the system is calculated as;

ΔU = -Q - W

ΔU = - 5.97 x 10⁵ J  -  6.53 x 10⁴ J

ΔU = -6.623 x 10⁵ J

The magnitude of the change in internal energy = 6.623 x 10⁵ J

(b) the efficiency of the athlete;

[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]

Answer:

me...:(

Explanation:

Answer:

hello I'm online here thanks for the points (◔‿◔)

If two objects are the SAME temperature and are physically touching,

then

. . .

. . .

. . .

The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.

Answer:

The new force becomes (1/9)th of the original force.

Explanation:

The gravitational force between two masses is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

Where

r is the distance between masses,

If the new distance is, r' = 3r

The new force is given by :

[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(3r)^2}\\\\F'=\dfrac{1}{9}\times G\dfrac{m_1m_2}{r^2}\\\\F'=\dfrac{F}{9}[/tex]

So, the new force becomes (1/9)th of the original force.

Answer: [tex]56.4\ m/s^2, 201.42\ m/s^2[/tex]

Explanation:

Given

Rocket attain a velocity of [tex]v=282\ m/s[/tex] in a time period of [tex]t=5\ s[/tex]

It was brought jarringly back to rest in only [tex]t'=1.4\ s[/tex]

Acceleration is the change in velocity of the object over a period of time

(a) Acceleration in his direction of motion

[tex]\Rightarrow a=\dfrac{v-0}{t}\\\\\Rightarrow a=\dfrac{282}{5}\\\\\Rightarrow a=56.4\ m/s^2[/tex]

(b) acceleration opposite to his direction of motion i.e. deceleration is

[tex]\Rightarrow a_d=\dfrac{0-v}{t'}\\\\\Rightarrow a_d=\dfrac{-282}{1.4}\\\\\Rightarrow a_d=-201.42 \ m/s^2\\\Rightarrow a_d=201.42\ \text{decelration}[/tex]

Answer:

kinetic energy of the train = 2,910.6 x 10⁷ joule

Explanation:

Given:

Mass of train = 3.3 x 10⁷ kg

Speed of train = 42 m/s

Find:

kinetic energy of the train

Computation:

kinetic energy = (1/2)(m)(v²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)

kinetic energy of the train = (3.3 x 10⁷)(882)

kinetic energy of the train = 2,910.6 x 10⁷ joule

Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].

Explanation:

Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]

Speed = 42 m/s

Kinetic energy is the energy acquired by an object due to its motion.

Formula to calculate kinetic energy is as follows.

[tex]K.E = \frac{1}{2}mv^{2}[/tex]

where,

m = mass of object

v = speed of object

Substitute the values into above formula as follows.

[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]

Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].

speed = 40 m/s

Explanation:

Since the object is dropped, V0y = 0.

Vy = V0y - gt

= -(10 m/s^2)(4 s)

= -40 m/s

This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.

The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,

By using the first equation of motion,

v = u + at

initial velocity(u) = 0 m/s

acceleration(a) = 10 m/s²

v = 0 + 10×4

v = 40 meters/seconds

Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.

Learn more about equations of motion from here,

brainly.com/question/5955789

#SPJ2

Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

Answer:

The correct solution is "14.6875 kg".

Explanation:

Given values:

Force,

F = 47.0 N

Acceleration,

a = 3.20 m/s²

Now,

⇒ [tex]Force=Mass\times Acceleration[/tex]

or,

⇒       [tex]F=ma[/tex]

⇒    [tex]47.0=m\times 3.20[/tex]

⇒       [tex]m=\frac{47.0}{3.20}[/tex]

⇒           [tex]=14.6875 \ kg[/tex]

Answer: [tex]4.65\ m/s[/tex]

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

[tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]

So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]

Answer: The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.

Explanation:

Quickly place the egg over the mouth of the bottle. The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

Answer:

T = 6.43 x 10⁻⁵ N.m

Explanation:

First, we will calculate the deceleration of the apparatus by using the third equation of motion:

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

where,

α = angular decelration = ?

θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad

ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s

ωf = final angular speed = 0 rad/s

Therefore,

[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]

negative sign shows deceleration

Now, for torque:

T = Iα

where,

T = Torque = ?

I = moment of inertia = 0.076 kg.m²

Therefore,

T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)

T = 6.43 x 10⁻⁵ N.m

Force is mass into acceleration

and pressure is force applied per unit area.

Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.

In Non-Impact printers, no mechanical moving component is used.

mark me brainliesttt :)))

Answer:

Option D

Explanation:

From the question we are told that:

The attractive magnetic force per unit length as

 [tex]f = F/L[/tex]

Separation Distance [tex]x=2d[/tex]

Generally the equation for  Magnetic force between two current carrying wire is mathematically given by

[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]

[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]

Where

[tex]x=2r[/tex]

And

[tex]I_1=I_2=>2I[/tex]

Then

[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]

[tex]\frac{F}{\triangle l}=>2f[/tex]

Therefore s the force per unit length between the wires if their separation is 2d

[tex]\frac{F}{\triangle l}=>2f[/tex]

Option D

Complete question:

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.

Answer:

The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

Explanation:

Given;

power of the flux produced, P = 1 W = 1 J/s

energy per photon, E = 78 keV

Convert the energy per photon to J

E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon

let the number of photons = n

n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s

[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]

Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

Answer: [tex]14.64\ N[/tex]

Explanation:

Given

Inclination of ramp is [tex]\theta=15^{\circ}[/tex]

Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal

Weight of cart [tex]W=40\ lb[/tex]

from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp

Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position

[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]

Answer:

the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

Explanation:

The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

            F = q E

You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest

Answer:

4.408 [tex]\mathsf{M_{sun}}[/tex]

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]

combined mass (m₁+m₂)  = 4.408 [tex]\mathsf{M_{sun}}[/tex]

Answer:

B) Inertia is the resistance of any physical object

Answer:

D

Explanation:

Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0

Answer:

[tex](b)\ t_1 - t_0[/tex]

[tex](d)\ t_2 - t_1[/tex]

[tex](e)\ \frac{t_2 - t_0}{2}[/tex]

Explanation:

Given

See attachment for complete question

Required

How long to reach the ground from the maximum height

First, calculate the time of flight (T)

[tex]T =t_2 - t_0[/tex]

The time taken (t) from maximum height to the ground is:

[tex]t = \frac{1}{2}T[/tex]

So, we have:

[tex]t = \frac{t_2 - t_0}{2}[/tex]

Another representation is:

At ymax, the time is: t1

On the ground, the time is t2

The difference between these times is the time taken.

So;

[tex]t = t_2 - t_1[/tex]

Since air resistance is to be ignored, then

[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height

Answer:

Resistance is as large as 2.8 ohm

Explanation:

Complete question

A 3.0 A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance?

Solution -

The relation between Power and current is as follows  

P = I^2*R

R = P/I^2

Were P = Power

R = resistance and

I = current

Given-  

P = 25 W  

I = 3.0 A

Substituting the given values, in above equation, we get -  

R = 25/3.0^2

R = 2.8 ohm