For a standard production car, the highest roadtested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution went from zero to 26.8 m/s (60 mph) in 3.275 s. Find the magnitude of the car's acceleration.

Answer:

[tex]Q = \frac{0.068}{E}[/tex]

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

[tex]QE = F_x[/tex]

[tex]mg = F_y[/tex]

since string makes 30 degree angle with the vertical so we will have

[tex]tan\theta = \frac{F_x}{F_y}[/tex]

[tex]tan30 = \frac{QE}{mg}[/tex]

[tex]Q = \frac{mg}{E}tan30[/tex]

[tex]Q = \frac{0.012\times 9.81}{E} tan30[/tex]

[tex]Q = \frac{0.068}{E}[/tex]

where E = electric field intensity

The net charge on the small plastic ball which is suspended by a string in a uniform is 0.068/E C.

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

A point charge on a string in equilibrium with an electric field can be given as,

[tex]E=\dfrac{mg \tan\theta}{Q}[/tex]

Here, (m) is the mass, (g) is acceleration due to gravity, and (Q) is the charge.

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field.

The ball is in equilibrium when the string makes a 30 degrees angle with the vertical. Thus, by the above formula,

[tex]E=\dfrac{(0.012)(9.81)\tan(30)}{Q}\\Q=\dfrac{0.068}{E}\rm\; C[/tex]

Thus, the net charge on the small plastic ball which is suspended by a string in a uniform is 0.068/E C.

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Answer:

  a.  17.0 m²

Explanation:

The product of the two dimensions is 16.99013 m². The least precise contributor has 3 significant figures, so the most precise result available is one with 3 significant figures: 17.0 m².

__

Additional comment

Given that each measurement may be in error by 1/2 of a least-significant digit, their product can be as little as 16.9700025, or as great as 17.0102625. This amounts to 16.9901325 ± 0.0201300. The value 17.0 suggests a range from 16.95 to 17.05, which exceeds the actual range possible with the given measurements. On the other hand, a 4 significant-figure value (16.99) suggests a much smaller range in the product than there may actually be: (16.985, 16.995)

Answer:hh

The acceleration may point in the opposite direct as the velocity

Explanation:

By definition free-fall is an uniform accelerate movement so speed always is increasing in fact acelertion and velocity always point in the same direction

Answer:

T h e  a c c e l .  e r a t i o n   m a y   p o i n t   i n   t h e   o p p o s i t e   d i r e c t   a s   t h e   v e l o c i t y   .  

Explanation:

... prejudice based on misogyny.

Also a good bit of stupidity.

The electrostatic constant is also known as Coulomb's constant or the electric constant. Its unit is [tex]\rm \frac{Nm^2}{C^2}[/tex].

It is a fundamental physical constant that appears in Coulomb's law, which describes the interaction between electric charges. It's denoted by the symbol K.

Given information:

Unit of force (F) = Newton

Unit of charge (Q) = Coloumb

Unit of distance (r) = Meters

Now,

On substituting the values of given units:

[tex]\rm F = \frac{Kq_1q_2}{r^2}\\N =\frac{K C\times C}{m^2}\\N = \frac{K C^2}{m^2}\\K = \frac{Nm^2}{C^2}[/tex]

The unit of electrostatic constant (K) is [tex]\rm \frac{Nm^2}{C^2}[/tex]. This constant (K) plays a crucial role in understanding and calculating the forces and interactions between charged particles in the realm of electrostatics.

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Final answer:

The unit for the electrostatic constant k is Newton meters squared per Coulomb squared (N·m²/C²), and k is also known as Coulomb's constant with a value of approximately 8.99 × 10⁹ N·m²/C².

Explanation:

The unit for the electrostatic constant k in the formula F = k[tex]q_{1}[/tex][tex]q_{2}[/tex] /r² can be determined by rearranging the formula to solve for k, which gives us k = Fr² / [tex]q_{1}[/tex][tex]q_{2}[/tex] . Since the unit for the electrostatic force F is Newtons (N), the distance r is in meters (m), and the charge [tex]q_{1}[/tex]and [tex]q_{2}[/tex] are in Coulombs (C), the units for k would be Newton meters squared per Coulomb squared (N·m²/C²). The value of the electrostatic constant k is approximately 8.99 × 10⁹ N·m²/C², also known as Coulomb's constant.

Answer:

Technician A says that this is because the front wheel cylinders are closer to the master cylinder than are the rear wheel cylinders.

Explanation:

Since the position of cylinder is near the front wheel so the normal force of the front wheel is more than the normal force on the rear wheel

as we know that the center of mass of the wheel is shifted towards the front wheel as the balancing is done with reference to its center of mass

so we will have

[tex]N_1 d_1 = N_2 d_2[/tex]

also we have

[tex]N_1 + N_2 = W[/tex]

so here we can say

friction force on front wheel will be more

[tex]F_f = \mu N_1[/tex]

so here front wheel has to do more work to stop the vehicle

Technician B is correct; the front brakes do more work because the vehicle's weight shifts to the front during braking due to dynamic load transfer. This is a physics principle related to force and motion, rather than the hydraulic pressure distribution which is equal for all brakes due to Pascal's principle.

Technician B suggests that the front brakes do more work than the rear brakes because the weight of a vehicle shifts to the front during a stop. This is correct, as deceleration causes the vehicle's weight to transfer to the front due to inertia, resulting in greater pressure on the front brakes. This phenomenon is known as weight transfer or dynamic load transfer. The claim by Technician A that proximity to the master cylinder affects braking force is not accurate, as hydraulic systems utilize Pascal's principle to ensure equal pressure distribution throughout the brake fluid regardless of the distance from the master cylinder.

Final answer:

After accounting for a construction zone, the minimum speed needed for the rest of the journey to arrive in time for the interview is approximately 132.67 km/h to cover the remaining 137 km in about 1 hour and 2 minutes.

Explanation:

The question involves calculating the least speed needed for the rest of the trip to arrive in time for an interview after encountering a construction zone. Given that the total distance is 300 km and the interview is at 11:15 am, with a departure time of 8:00 am, we need to calculate the travel time at two different speeds and the remaining distance and time available.

First, let's calculate the time taken for the first part of the journey at 100 km/h for 120 km:
Time = Distance / Speed
Time = 120 km / 100 km/h = 1.2 hours.

Next, for the construction zone at 42 km/h for 43 km:
Time = 43 km / 42 km/h ≈ 1.024 hours.

The total time taken so far is 1.2 + 1.024 ≈ 2.224 hours. Since the student left at 8:00 am, by this time it would be approximately 10:13 am (8:00 am plus approximately 2 hours and 13 minutes).

With the interview at 11:15 am, there's 1.0333 hours (1 hour and 2 minutes) left to travel the remaining distance of 300 km - 120 km - 43 km = 137 km.

To find the minimum speed needed for the remaining distance, we use the formula: Speed = Distance / Time Required
Minimum Speed = 137 km / 1.0333 hours ≈ 132.67 km/h.

The least speed needed for the remainder of the trip is approximately 132.67 km/h.

Answer:

Part a)

[tex]\Delta x = 36.35 m[/tex]

Part b)

height will be 107.2 m

Part c)

[tex]t = 1.78 s[/tex]

Explanation:

Initial velocity of the ball is given as

v = 7.90 m/s

angle of projection of the ball

[tex]\theta = 23^o[/tex]

now the two components of velocity of ball is given as

[tex]v_x = 7.90 cos23 = 7.27 m/s[/tex]

[tex]v_y = 7.90 sin23 = 3.09 m/s[/tex]

Part a)

Since ball strike the ground after t = 5 s

so the distance moved by the ball in horizontal direction is given as

[tex]\Delta x = v_x \times t[/tex]

[tex]\Delta x = 7.27 (5)[/tex]

[tex]\Delta x = 36.35 m[/tex]

Part b)

Now in order to find the height of the ball we can find the vertical displacement of the ball

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]\Delta y = (3.09)(5) - \frac{1}{2}(9.81)(5^2)[/tex]

[tex]\Delta y = -107.2 m[/tex]

So height will be 107.2 m

Part c)

when ball reaches a point 10 m below the level of launching then the displacement of the ball is given as

[tex]\Delta y = -10 m[/tex]

so we will have

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]-10 = 3.09 t - \frac{1}{2}(9.81)t^2[/tex]

so we will have

[tex]t = 1.78 s[/tex]

Answer:

v = -431.2 m/s

Explanation:

Given that,

Initial position of the object, [tex]x=640-4.9t^2[/tex]

Let v is its speed 44 second after it is dropped. The relation between the speed and the position is given by :

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(640-4.9t^2)}{dt}[/tex]

[tex]v=-9.8t[/tex]

Put t = 44 seconds in above equation. So,

[tex]v=-9.8\times 44[/tex]

v = -431.2 m/s

So, the speed of the ball 44 seconds after it is dropped is 431.2 m/s and it is in moving downwards.

Final answer:

The speed of the object 44 seconds after being dropped from a 640-meter-high cliff is 431.2 m/s. This is found by differentiating the height function to get the velocity function and substituting the time into the velocity equation.

Explanation:

To determine the speed of an object 44 seconds after it is dropped from the top of a 640-meter-high cliff, you can differentiate the height function to find the velocity function.

Given the height formula h(t) = 640 - 4.9t², we get the velocity function v(t) = h'(t) = -9.8t by differentiation. At t = 44 seconds, the speed is simply the absolute value of the velocity. So, we calculate: v(44) = -9.8 * 44

This yields a velocity of -431.2 m/s, and since speed is the magnitude of velocity, the speed is 431.2 m/s.

Answer:

25%

Explanation:

If there is 20% scrap, then

80% of (production + scrap) = 72 pcs/hr

Now,

[tex]\frac{80}{100} \times (72+scrap)= 72 pcs/hr\\0.80\times (72+Scrap) = 72 \\ 72+Scrap = \frac{72}{0.80}\\72 + Scrap = 90\\Scrap = 18 pcs /hr[/tex]

The percent increase in labor productivity is, [tex]\frac{18}{72} =0.25[/tex].

Or it can be written as  25% of  which would give 80 pieces per hour.                                            

Answer:

A. carrying a box of crayons across the room.

Explanation:

work is said to be done when a force moves something over a distance.

Answer:

A.

Explanation:

Carrying a box of crayons across the room.

Answer: a, d

Explanation:

A- eaven if the direction changes it's still 1D

B- an airplane always needs more than 1D for take off and landing

C- it isn't likely to get there by an straight way without any land relief

D- a car in a straight road its 1D if the load has no land relief.

Answer:

The answer to your question is: mass = 38.93 kg

Explanation:

Data

mass = ?

Weight = 382 N

gravity = 9.81 m/s2

Formula

Weight = mass x gravity

mass = weight / gravity

mass = 382 / 9.81         substitution

mass = 38.93 kg           result

The mass of a dog that weighs 382 N can be calculated using the weight formula w=mg, rearranged as m=w/g. Substituting the values, we find that the mass of the dog is approximately 39 kg.

To calculate the mass of a dog that weighs 382 N, we use the formula w = mg, where w is weight, m is mass, and g is the acceleration due to gravity, which on Earth is approximately 9.8 m/s². So, to find the mass, we rearrange the formula to m = w/g.

Substituting the given values into the equation, m = 382 N / 9.8 m/s² = 39.0 kg. So, the mass of the dog is approximately 39 kg when rounded to the nearest whole number.

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Answer:

a) heat flows from the warm plate to the cold one

Explanation:

Heat is an expression of energy, the bodies with more heat mean that they have more energy.  In this way, they can share or pass this "excess" of energy to the cold body. Both plates are not in total contact but some part are close enough (they are touching) that the heat can pass from one to the other. Also the heat always need a media to travel, in this case, the air that surrounds the plates is also the media that the heat travels from the hot to the cold one.

Answer:

- 3.33 m /s

Explanation:

m1 = 5 kg

m2 = 10 kg

u1 = 10 m/s

u2 = - 10 m/s

Let the velocity of the combination of the blob is v.

Use the conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v[/tex]

5 x 10 - 10 x 10 = (5 + 10)v

50 - 100 = 15 v

v = - 3.33 m /s

Thus, the velocity of the blob after sticking together is - 3.33 m/s.

Answer:

No, it is impossible

Explanation:

Kinematics equation:

[tex]Vf^{2} =Vo^{2} -2gy[/tex]

if height is maximum:

y=H and Vf=0

so:

Analysis: From the last equation we see that the maximum height depends ONLY on the initial speed. This means that if both objects reach the same maximum height, then they necessarily need to have the SAME initial velocity. If they have the same initial velocity and in order to reach the maximum height at the SAME time the only way is that they are released at the SAME TIME.

Answer:

Acceleration, [tex]a=2.48\ m/s^2[/tex]

Explanation:

Given that,

The plane is at rest initially, u = 0

Final speed of the plane, v = 72.2 m/s

Time, t = 29 s

We need to find the average acceleration for the plane. It can be calculated as :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{72.2}{29}[/tex]

[tex]a=2.48\ m/s^2[/tex]

So, the average acceleration for the plane is [tex]2.48\ m/s^2[/tex]. Hence, this is the required solution.

Answer:

F = 50 N

Explanation:

As we know by Newton's II law

{tex]F = ma[/tex]

force is product of mass and acceleration

so we will have

[tex]m = 500 kg[/tex]

[tex]a = 10 cm/s^2[/tex]

[tex]a = 0.10 m/s^2[/tex]

now we have

[tex]F = 500 \times 0.10 [/tex]

[tex]F = 50 N[/tex]

The magnitude of the rightward net force acting on the object is 50 N.

Force can be defined as the product of mass and acceleration.

To calculate the magnitude of the rightward net force, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The magnitude of the rightward net force acting on the object is 50 N.

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Answer:

Option a

Explanation:

The three basic and major points of the kinetic theory of gases are listed below:

1. In the collision between the gas molecules, no loss or gain of energy takes place.

2. According to the theory, the gaseous molecules exhibits constant and linear motion.

3. The molecules of the gas occupies very little space as compared to the container. Thus the empty space in a gas is not very little.

Answer:

The option that does NOT correspond to the kinetic theory of gases is the option a."There is very little empty space in a gas."

Explanation:

The kinetic theory of gases allows to deduce the properties of the ideal gas using a model in which the gas molecules are spheres that comply with the laws of classical mechanics. This theory states that heat and movement are related, that particles of all matter are moving to some extent and that heat is a sign of this movement.

The postulates of this theory are:

Taking into account the above, the option that does NOT correspond to the kinetic theory of gases is the option a."There is very little empty space in a gas."

Answer:

The distance travelled is 151.22m and it took 0.97s

Explanation:

Well, this is an ARM problem, so we will need the following formulas

[tex]x(t)=x_{0} +v_{0} *(t-t_{0} )+0.5*a*(t-t_{0} )^{2}[/tex]

[tex]v(t)=v_{0} +a*(t-t_{0} )[/tex]

where [tex]x_{0}[/tex] is the initial position (we can assume is zero), [tex]v_{0}[/tex] is the initial speed of 104 m/s, [tex]t_{0}[/tex] is the initial time (we also assume is zero), a is the acceleration of 107 m/s2, v is speed, x is position and t is time.

Now that we have the formulas, we know that when the electron stops it has no speed. Then we calculate how much time it takes to stop.

[tex]0=104m/s-107m/s^{2} *t\\t=0.97s[/tex]

Finally, we calculate the distance travelled in this time

[tex]x(0.97s)=104m/s*0.97s+0.5*107m/s^{2}*(0.97s)^{2}=151.22m[/tex]

Answer:

Part a)

[tex]d = 5 m[/tex]

Part b)

[tex]T = 2\times 10^{-3} s[/tex]

Explanation:

Part a)

The electron will move in this forbidden region till its speed will become zero

So here we will have

[tex]v_f = 0[/tex]

[tex]v_i = 10^4 m/s[/tex]

also its deceleration is given as

[tex]a = - 10^7 m/s^2[/tex]

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - (10^4)^2 = 2(-10^7) d[/tex]

[tex]d = 5 m[/tex]

Part b)

Now the time till its speed is zero

[tex]v_f - v_i = at[/tex]

[tex]0 - 10^4 = -10^7 t[/tex]

[tex]t = 10^{-3} s[/tex]

so total time that it will be in the region is given as

[tex]T = 2 t[/tex]

[tex]T = 2\times 10^{-3} s[/tex]

Answer:

7.9[tex]\frac{gr}{cm^3}[/tex]

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7[tex]cm^3[/tex]).

Density is [tex]\frac{mass}{volume}[/tex].

So the density of that piece of metal is [tex]\frac{55.3g}{7cm^3}[/tex]

Which leaves us with a final density of 7.9[tex]\frac{gr}{cm^3}[/tex]

Answer:

True.

Explanation:

He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater.  Hence you will be drown inside the water.  

Answer:

The acceleration of the object is 9.3 m/s²

Explanation:

For a straight movement with constant acceleration, this equation for the position applies:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:

x₁ = x0 + v0 t + 1/2 a (1 s)²

x₂ = x0 + v0 t + 1/2 a (2 s)²

x₂ - x₁ = 14 m

we know that the object starts from rest, so v0 = 0

substracting both equations of position we will get:

x₂ - x₁ = 14

x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m

x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m

2 a s² - 1/2 a s² = 14 m

3/2 a s² = 14 m  

a = 14 m / (3/2 s²) = 9.3 m/s²

The constant acceleration of the object is 9.33 m/s².

The given parameters;

The acceleration of the object is calculated by applying the second kinematic equation as follows;

s = ut  +  ¹/₂at²

where;

at t = 1.0 s

x₁ = 0  +   ¹/₂(1²)a

x₁ =  ¹/₂(a)

at t = 2.0 s

x₂ = 0 +  ¹/₂(2²)a

x₂ = 2a

The change in position;

Δx = x₂ - x₁ = 14

14 = 2a -  ¹/₂(a)

[tex]14 = \frac{3a}{2} \\\\3a = 2(14)\\\\3a = 28\\\\a = \frac{28}{3} = 9.33 \ m/s^2[/tex]

Thus, the constant acceleration of the object is 9.33 m/s².

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Answer:

D = 6.74 m

Explanation:

Let the f newtons  be the braking force

distance to stop =  x meters

speed =  v m/s

we know that work done is given as

work done W = fx joules

[tex]fx = \frac{1}{2} mv^2[/tex]

If the speed is doubled ,

[tex]fx' = \frac{1}{2} m(2v)^2[/tex]

  [tex] = 4[\frac{1}{2}mv^2] = 4fx[/tex]

stooping distance is D = 4x

[tex]40km/h = \frac{40*1000}{60*60} = 11.11 m/s[/tex]

60 km/hr = 16.66 m/s  

braking force = f

[tex]f*3 =  [\frac{1}{2}mv^2] = [\frac{1}{2}m*11.11^2][/tex]

[tex]f*D = [\frac{1}{2}m*16.66^2][/tex]

[tex]\frac{D}{3} =  \frac{16.66^2}{11.11^2}[/tex]

[tex]\frac{D}{3} = 2.2489[/tex]

D = 6.74 m

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

[tex]V=X/T[/tex]

First stage:

T1=5s

[tex]v_{f}  =v_{o} - at[/tex]

But, [tex]v_{f}  =0[/tex]   (decelerates to rest)

then: [tex]a =v_{o} /t=0.3/5=0.06m/s^{2}[/tex]

on the other hand:

[tex]x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m[/tex]

X1=75cm

Second stage:

T2=5s

[tex]x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m[/tex]

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

The average speed of the particle is 20 centimetre per second.

Average speed is given as total distance covered by the particle divided by the total time of the journey.

The particle decelerates from 30 cm/s to rest (0 cm/s) in 5.0 s. Hence, the deceleration can be given as,

a = (0 - 30 cm/s) / 5 s = - 6 cm/s²

From the kinematic equation:

s = ut + [tex]\frac{1}{2}at^2[/tex], where s = displacement, a = acceleration, t = time, and u = initial velocity of the particle, we can determine the displacement.

so, s = (30 cm/s) (5 s) - [tex]\frac{1}{2} (6 cm/s^2)(5 s)^2[/tex] = 150 cm - 75 cm = 75 cm

The particle then accelerates uniformly at 10 cm/s² for another 5.0 s. The final velocity after this period is given as:

v = u + at

Here, u = 0 cm/s, a = 10 cm/s². t = 5s. Hence,

v = (10 × 5) m/s = 50 cm/s

Using the kinematic equation:

v² = u² + 2as, we get:

(50 cm/s)² = 2 (10 cm/s²) s

or, s = 125 cm

Total displacement of the particle = 75 cm + 125 cm = 200 cm

total time of journey = 5s + 5s = 10s

so, average velocity = 200 cm / 10 s = 20 cm/s

Answer:

yes

Explanation:

because when you slow down, the resistance slows with the speed.

Answer:

Yes, air resistance decrease with speed

Explanation:

Air resistance is a kind of fluid friction that acts when objects flow through fluids. It is affected by the velocity of moving objects and the area of the objects. When an object moves with a greater velocity the air resistance acting on them will be high, so the speed decreases.

Fluid friction acts in fluids namely liquids and gases. In liquids the friction is called buoyancy and in gases it is called air resistance or drag. When two objects having the same mass but different area move through air, the object with larger area will have less velocity compared with the other object.

Explanation:

Given that,

Electrostatic force, [tex]F=3.7\times 10^{-9}\ N[/tex]

Distance, [tex]r=5\ A=5\times 10^{-10}\ m[/tex]

(a) [tex]F=\dfrac{kq^2}{r^2}[/tex], q is the charge on the ion              

[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]

[tex]q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}[/tex]      

[tex]q=3.2\times 10^{-19}\ C[/tex]

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

[tex]n=\dfrac{q}{e}[/tex]

[tex]n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}[/tex]

n = 2

Hence, this is the required solution.                        

Answer:

35 meters

Explanation:

The river flows at 2 m/s from North to South and motor can propel the boat at 5m/s. As the downstream direction is positive, we are considering the river flow also propels the boat adding its speed to the boat. It means the boat and the person in the inner tube are in fact moving at 7 m/s. Distance can be calculated as follows:

[tex]v = d/t[/tex]  ⇒[tex]d = vt[/tex] ⇒[tex]d = 7\frac{m}{s}x5s[/tex]

[tex]d = 35m[/tex]

The motorboat will cover a distance of 35 meters downstream in 5 seconds. This is calculated by adding the boat's velocity relative to the water (5 m/s) to the velocity of the river current (2 m/s), giving a resultant velocity of 7 m/s and multiplying it by the time interval.

We need to calculate the distance the motorboat will cover downstream in a 5-second interval. The speed of the river current is 2 m/s and the speed of the motorboat relative to the water is 5 m/s. Since we are considering the downstream direction as positive, the boat's velocity relative to an observer on the shore would be the sum of these two speeds.

The boat's speed relative to the shore (resultant velocity) is:

Velocity of the boat relative to the shore = Velocity of the boat relative to the water + Velocity of the river = 5 m/s + 2 m/s = 7 m/s

The distance covered by the boat over a period of 5 seconds would be:
Distance = Velocity × Time = 7 m/s × 5 s = 35 m

Therefore, the motorboat will move 35 meters downstream as observed by someone outside of the water, like the person in the inner tube.

Answer:

Potential Energy

Explanation:

The electron transport chain is a series of proteins and organic molecules located in the inner membrane of the mitochondria.During electron transfer and proton pumping electrons are moved from a higher energy level to a lower one and in the process release energy.This energy is used to make ATP.It is stored in the electrochemical gradient of protons and it has to be released for electron transport to continue.

Creation of a proton gradient by the electron transport chain represents a pivotal process in cell respiration known as oxidative phosphorylation, in which the energy from electrons is used to form a gradient of protons. This gradient powers ATP production, converting the energy of electrons to a form suitable for cellular work.

The creation of a proton gradient by the electron transport chain represents a crucial step in cell respiration called oxidative phosphorylation. In the mitochondria, the electron transport chain uses energy from electrons to pump protons (hydrogen ions) from the mitochondrial matrix into the intermembrane space, forming a gradient. This proton gradient is the potential energy that drives ATP production when the protons flow back across the membrane via ATP synthase. Therefore, the electron transport chain converts the energy of electrons into a useable form for the cell to perform work.

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Water leaves a fireman’s hose (held near the ground) with an initial velocity v0= 22.5 m/s at an angle θ = 28.5° above horizontal, the horizontal distance d from the building base where the fireman should place the hose for the water to reach its maximum height as it strikes the building is given by [tex]\( \frac{v_0^2 \cdot \sin(2\theta)}{g} \)[/tex]

a) To find the time [tex]\( t_{\text{max}} \)[/tex] that the water travels to reach its maximum vertical height, we can use the following kinematic equation for vertical motion:

[tex]\[ v_y = v_{0y} - g \cdot t \][/tex]

Where:

[tex]\( v_y \)[/tex] is the vertical component of velocity at time t.

[tex]\( v_{0y} \)[/tex] is the initial vertical component of velocity (which is [tex]\( v_0 \cdot \sin(\theta) \)[/tex])

g is the acceleration due to gravity

At the maximum height, the vertical component of velocity becomes zero, so we can set [tex]\( v_y = 0 \)[/tex] and solve for [tex]\( t_{\text{max}} \)[/tex]:

[tex]\[ 0 = v_{0y} - g \cdot t_{\text{max}} \][/tex]

[tex]\[ t_{\text{max}} = \frac{v_{0y}}{g} \][/tex]

Substituting [tex]\( v_{0y} = v_0 \cdot \sin(\theta) \)[/tex], we get:

[tex]\[ t_{\text{max}} = \frac{v_0 \cdot \sin(\theta)}{g} \][/tex]

b) To determine the horizontal distance d from the building base where the fireman should place the hose for the water to reach its maximum height as it strikes the building, we need to consider the horizontal motion of the water.

The horizontal distance d can be calculated using the following equation for horizontal motion:

[tex]\[ d = v_{0x} \cdot t_{\text{max}} \][/tex]

Where:

[tex]\( v_{0x} \)[/tex] is the initial horizontal component of velocity (which is [tex]\( v_0 \cdot \cos(\theta) \)[/tex])

[tex]\( t_{\text{max}} \)[/tex] is the time it takes to reach the maximum height (calculated in part a)

Substituting [tex]\( v_{0x} = v_0 \cdot \cos(\theta) \)[/tex] and [tex]\( t_{\text{max}} = \frac{v_0 \cdot \sin(\theta)}{g} \)[/tex], we get:

[tex]\[ d = (v_0 \cdot \cos(\theta)) \cdot \left(\frac{v_0 \cdot \sin(\theta)}{g}\right) \][/tex]

Simplifying:

[tex]\[ d = \frac{v_0^2 \cdot \sin(\theta) \cdot \cos(\theta)}{g} \][/tex]

This can be further simplified using the trigonometric identity [tex]\( \sin(2\theta) = 2 \sin(\theta) \cdot \cos(\theta) \)[/tex]:

[tex]\[ d = \frac{v_0^2 \cdot \sin(2\theta)}{g} \][/tex]

Thus, the horizontal distance d from the building base where the fireman should place the hose for the water to reach its maximum height as it strikes the building is given by [tex]\( \frac{v_0^2 \cdot \sin(2\theta)}{g} \)[/tex].

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The time, tmax, the water takes to reach its maximum vertical height can be calculated using the formula tmax = v0sin(θ) / g. The horizontal distance, d, from the building base where the fireman should place the hose can be found using the formula d = v0cos(θ)tmax.

To find the time, tmax, the water travels to reach its maximum vertical height, we can use the formulatmax = v0sin(θ) / g

Where v0 is the initial velocity (22.5 m/s), θ is the angle above horizontal (28.5°), and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the values into the formula, we get:

tmax = (22.5 m/s)sin(28.5°) / 9.8 m/s^2

Solving this equation will give us the value for tmax.

To determine the horizontal distance, d, from the building base where the fireman should place the hose, we can use the formula:

d = v0cos(θ)tmax

Where v0 is the initial velocity, θ is the angle above horizontal, and tmax is the time calculated in part (a).

Substituting the values into the formula, we get:

d = (22.5 m/s)cos(28.5°)tmax

This equation will give us the value for d in terms of v0, θ, and g.