1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) the time taken for it to return to 0 (c) the taken for it to be 35m below 0

[tex]{\large{\bold{\rm{\underline{Given \; that}}}}}[/tex]

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

[tex]{\large{\bold{\rm{\underline{To\; find}}}}}[/tex]

★ The speed of the hound and the hare

[tex]{\large{\bold{\rm{\underline{Solution}}}}}[/tex]

★ The speed of the hound and the hare = 25:18

[tex]{\large{\bold{\rm{\underline{Full \; Solution}}}}}[/tex]

[tex]\dashrightarrow[/tex]  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

 But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

 Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

 Now the distance travelled by the hare in it's 6 leaps..!

 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

Answer:

[tex]E=6.8Kv/m[/tex]

Explanation:

From the question we are told that

Distance b/w plate [tex]d=10cm=>0.1m[/tex]

P_1 Potential at 7.35 [tex]V=533v[/tex]

Generally the equation for electric field at a distance is mathematically given as

[tex]E=\frac{v}{d}[/tex]

[tex]E=\frac{533}{7.85*10^-^2}[/tex]

[tex]E=6789.808917[/tex]

[tex]E=6.8*10^3[/tex]

[tex]E=6.8Kv/m[/tex]

Answer: take less time to go the same distance

Explanation:

Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.

Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.

Elements: Nitrogen; Oxygen; Phosphorus; Selenium...

Answer:

70 kg

Explanation:

divide it by 2

Hope this helped!

Answer:

63.502932 Kilograms

Explanation:

I'm pretty sure I the third option C.

Explanation:

sorry if I'm wrong

Answer:

Displacement

Explanation:

The quantity 45m north is a typical example of displacement.

Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

Answer:

See explanation below

Explanation:

Psychoactive drugs are drugs that affect the central nervous system. They alter cognitive function by changing mood and consciousness.

Examples;

Depressants: Inhibit the function of the central nervous system and neural activity.

Stimulants: Excite neural activity and temporarily elevate awareness.

Amphetamines: Increase dopamine activity and produce schizophrenic-like symptoms.

Hallucinogens: Distort perceptions and effects on thinking.

A drug is any substance that alters how the body functions.

A drug is any substance that alters how the body functions. There are different types of drugs that affect different parts of the body.

We shall now explain the following classifications of drugs;

Learn more about drugs: https://brainly.com/question/6022349

Answer:

I think the bucs are gonna win because Tom Brady is on their team and it's rigged

but maybe I'm just thinking negatively lol

1. measure of how quickly velocity is changing . . . acceleration

2. speed in a given direction . . . velocity

3. force that resists moving one object against another . . . friction

4. measure of the pull of gravity on an object . . . weight

5. tendency of an object to resist a change in motion . . . inertia

6. size . . . magnitude

1. measure of how quickly velocity is changing is acceleration.

2. speed in a given direction is velocity.

3. force that resists moving one object against another is friction.

4. measure of the pull of gravity on an object is weight.

5. tendency of an object to resist a change in motion is inertia.

Acceleration has the term used in mechanics to describe the pace at which the velocity of an object varies over time. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it.

It is a vector quantity with an SI unit is m/s² and the dimension formula is LT⁻². A massive body will accelerate or alter its velocity at a constant rate when a constant force is applied to it, according to Newton's second law. In the simplest case, when a force is applied to an object at rest, it accelerates in the force's direction.

Therefore, 1. measure of how quickly velocity is changing is acceleration.

2. speed in a given direction is velocity.

3. force that resists moving one object against another is friction.

4. measure of the pull of gravity on an object is weight.

5. tendency of an object to resist a change in motion is inertia.

Learn more about friction on:

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#SPJ5

Answer:

I think it is heterogeneous mixture. have a good day

Answer:

heterogeneous mixture

Explanation:

i took the test

Answer:

Earth

lol... ....

Answer:

a.-7.5 kg m/s

b.-375 N

Explanation:

We are given that

Mass of football, m=0.5 kg

Initial velocity ,u=15m/s

Final speed ,v=0

Time, t=0.02 s

a. We have to find the impulse delivered to the ball.

We know that

Impulse=Change in momentum

I=m(v-u)

Using the formula

[tex]I=0.5(0-15)=-7.5 kg m/s[/tex]

Hence,  the impulse delivered to the ball=-7.5 kg m/s

(b)

We know that

Force,[tex]F=\frac{|mpulse}{time}[/tex]

Using the formula

[tex]F=\frac{-7.5}{0.02}=-375 N[/tex]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

[tex]s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}[/tex] (1)

Speeder

[tex]s = s_{o} + v_{o,S}\cdot t[/tex] (2)

Where:

[tex]s_{o}[/tex] - Initial position, measured in meters.

[tex]s[/tex] - Final position, measured in meters.

[tex]v_{o,P}[/tex], [tex]v_{o,S}[/tex] - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

[tex]a[/tex] - Acceleration of the unmarked police car, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t'[/tex] - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

[tex]v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t[/tex]

[tex]v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t[/tex]

[tex]\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0[/tex]

If we know that [tex]a = 1.6\,\frac{m}{s^{2}}[/tex], [tex]v_{o,P} = 0\,\frac{m}{s}[/tex], [tex]v_{o,S} = 53.4\,\frac{m}{s}[/tex] and [tex]t' = 2.2\,s[/tex], then we solve the resulting second order polynomial:

[tex]0.8\cdot t^{2}-56.92\cdot t +3.872 = 0[/tex] (3)

[tex]t_{1} \approx 71.082\,s[/tex], [tex]t_{2}\approx 0.068[/tex]

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Answer:

6.69 m/s

4.483 m

1.42s

Explanation:

Given that:

Initial Velocity, u = 0

Final velocity, v =?

Acceleration, a = 35m/s²

1.) using the relation :

v² = u² + 2as

v² = 0 + 2(35) * 64*10^-2m

v² = 70 * 0.64

v = sqrt(44.8)

v = 6.693

v = 6.69 m/s

B.) height from the ground, h0 = 2.2

How high ball went , h:

Using :

v² = u² + 2as

Upward motion, g = - ve

0 = 6.69² + 2(-9.8)*(h - 2.2)

0= 6.69² - 19.6(h - 2.2)

44.7561 + 43.12 - 19.6h = 0

19.6h = 44.7561 - 43.12

h = 87.8761 / 19.6

h = 4.483 m

C.)

vt - 0.5gt² = h - h0

6.69t - 0.5(9.8)t²

6.69t - 4.9t² = 1.83 - 2.2

-4.9t² + 6.69t + 0.37 = 0

Using the quadratic equation solver :

Taking the positive root:

1.4185 = 1.42s

Answer:

6.86 m/s

Explanation:

The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.

I suppose that the only force, in this case, is the gravitational force acting on the fish.

Then the gravitational equation of the fish will be:

a(t) = -9.8m/s^2

For the velocity equation we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial velocity of the fish and is what we want to find.

For the position equation we need to integrate over time again to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0

p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m

Then the equation is:

p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t

p(t) = (-4.9 m/s^2)*t^2 + v0*t

We know that the maximum height is 2.4m

The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:

v(t) = (-9.8m/s^2)*t + v0 = 0

      t = v0/9.8m/s^2

Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m

2.4m = p( v0/9.8m/s^2) =  (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)

2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))

2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)

2.4m = 0.5*v0^2/(9.8m/s^2)

2.4m/0.5 = v0^2/(9.8m/s^2)

4.8m*(9.8m/s^2) = v0^2

√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s

Answer:

A. Retina

Explanation:

Answer:

x = 1.04866

Explanation:

Force can be defined from power energy by the expressions

          F = [tex]- \frac{ dU}{ dx}[/tex]

in this case we are the expression of the potential energy

          U = [tex]\frac{2.6}{x^{8} } - \frac{4.3}{ x^{4} }[/tex]

let's find the derivative

         dU / dx = 2.6 ( [tex]\frac{-8}{x^{9} }[/tex]) - 4.3 ([tex]\frac{-4}{ x^{5} }[/tex])

         dU / dx = [tex]- \frac{20.8}{ x^{9} } + \frac{17.2 }{ x^{5} }[/tex]

we substitute

          F = + \frac{20.8}{ x^{9} }  - \frac{17.2 }{ x^{5} }

at the equilibrium point the force is zero, so

           [tex]\frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }[/tex]

           20.8 / 17.2 = x⁴

            x⁴ = 1.2093

             x = [tex]\sqrt[4]{ 1.2093}[/tex]

             x = 1.04866

Answer:

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Explanation:

Data Given:

Height = 25000 ft

Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s

[tex]m_{s} = 1.2m_{e}[/tex]

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

Where,

[tex]m_{s}[/tex] = Mass flow rate of fan

[tex]m_{e}[/tex] = Mass flow rate of core

F = Thrust

Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]

Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]

Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa

Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa

Thrust = F = 10580 lbf = 47062.2 N

Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s

Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s

At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]

Now,

we have:

[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x  [tex]D_{e}[/tex]  x [tex]A_{e}[/tex]

Plugging in the values, we get:

[tex]m_{e}[/tex]  = 168.16 [tex]A_{e}[/tex]   Equation 1

And,

[tex]m_{s}[/tex]  = [tex]D_{s}[/tex]  x [tex]A_{s}[/tex] x  [tex]u_{s}[/tex]

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]  Equation 2

As, we know,

[tex]m_{s} = 1.2m_{e}[/tex]  

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]

And now for Thrust, we have:

F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex]  - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex]  - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex]  -  [tex]u_{a}[/tex] ) Equation 3

Now, substitute equation 1 and 2 in equation 3, we get:

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate  / Thrust

TSFC = [tex]m_{f}[/tex]/F

And,

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

[tex]m_{e}[/tex]   =  60.94

[tex]m_{f}[/tex]  = 0.0255 x 60.94

[tex]m_{f}[/tex]  = 1.55397

TSFC = [tex]m_{f}[/tex]/F

TSFC = 1.55397/47062.2

TSFC = 3.301 x [tex]10^{-5}[/tex]

Low TSFC = High efficiency

High TSFC = Low efficiency

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Answer:

displacement is the distance between the starting point and the ending point of an object's journey

Answer:

25.3J

Explanation:

Given parameters:

Mass of aluminum  = 3.05g

Initial temperature  = 10.8 °C

Final temperature  = 20 °C

Specific heat  = 0.9J/g °C

Unknown:

Amount of heat needed for the temperature to change  = ?

Solution:

To solve this problem, we use the expression:

       H  = m C Ф  

H is the amount of heat

m is the mass

C is the specific heat capacity

Ф is the change in temperature

     H  = 3.05 x 0.901 x (20 - 10.8) = 25.3J

Answer:

The response to this question is as follows:

Explanation:

The whole question and answer can be identified in the file attached, please find it.

The force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.

The given parameters;

The force diagram of all the forces acting on the snowball is calculated as follows;

                                     ↑ N

                                     ⊕  → F

                                      ↓ W

Where;

Thus, the force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.

Learn more here:https://brainly.com/question/3624253

Acceleration = (change in velocity) / (time for the change)

Change in velocity = (ending velocity) - (starting velocity)

Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)

Change in the plane's velocity = 2,000 m/s north

Time for the change = 40 seconds

Acceleration = (2,000 m/s north) / (40 seconds)

Acceleration = 50 m/s² north

Answer:

[tex]\triangle U=-e (V_2-V_1)[/tex]

[tex]\triangle U=130eV[/tex]

[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]

Explanation:

From the question we are told that

The potential at point 1, [tex]V_1 = 24V[/tex]

The potential at point 2, [tex]V_2 = 154V[/tex]

a)Generally work done by proton is given as

 [tex]w=-\triangle U[/tex]

 [tex]e\triangle V=-\triangle U[/tex]

 [tex]\triangle U=-e (V_2-V_1)[/tex]  

Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as

 [tex]\triangle U=-e (V_2-V_1)[/tex]

b)Generally the electric potential energy in electron volts (eV). is mathematically given as

 [tex]\triangle U=-e (154-24)V[/tex]

 [tex]|\triangle U| =|-e (130)V|[/tex]

 [tex]\triangle U=130eV[/tex]

c) Generally according to the law of conservation of energy

[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]

[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]

[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]

[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]

Answer:

A

Explanation:

Corrected, it's 2) 200