A school bus is moving 26.8 m/s on
flat ground when it begins to
accelerate at 4.73 m/s². How much
time does it take to travel 95.1 m?
(Unit=s)
Watch your minus signs!
time (s)
Enter

Answer: 0.025 cm

Explanation:

The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm

The magnification of the lens is given by

m = v/u = h'/h

v/u = h'/ h

h' = ( v/u ) × h

h' = ( 2cm /-120cm ) × 1.5cm

h' = - 0.025cm

Therefore, the height of the image is 0.025 cm

An externality, sometimes called a spillover, occurs when an exchange between a buyer and seller has an impact on a third party who is not part of the exchange. Externalities can be positive or negative

Externalities refers to situations when the effect of production or consumption of goods and services imposes costs or benefits on others which are not reflected in the prices charged for the goods and services being provided.

Externalities Private markets offer an efficient way to put buyers and sellers together and determine what goods are produced, how they are produced, and who gets them. The principle that voluntary exchange benefits both buyers and sellers is a fundamental building block of the economic way of thinking. But what happens when a voluntary exchange affects a third party who is neither the buyer nor the seller?

Externalities can be negative or positive. If you hate country music, then having it waft into your house every night would be a negative externality. If you love country music, then what amounts to a series of free concerts would be a positive externality.

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Answer:

STATES OF MATTER The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.

Explanation:

The reference point that should be used to describe Melanie motion is the pond (option C).

Reference point is a particular point in space which is used as an endpoint to measure a distance from or chart a map from.

According to this question, Melanie gets out of her car at the park and walks 25 m to the trail entrance. She jogs around the trail until she reaches a pond, where she stops briefly.

However, she then continues to follow the trail around the pond. This suggests that the pond should serve as the reference point for Melanie's motion.

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Answer:

(B)  (the leaves will separate when subjected to a net charge)

The image that represents a gaining of a charge on the leaves of the electroscope by conduction is image B.

An electroscope is a device that detects the presence of an electric charge or ionizing radiation by suspending a pair of thin gold leaves from an electrical conductor that leads to the outside of an insulating container.

When an electrical charge is brought close to or in contact with a conductor, the leaves in an electroscope spread apart.

This occurs because the same charge, whether positive or negative, is received by the leaves, causing their repulsion (spreading apart).

A negatively charged rod is brought in close proximity to an uncharged electroscope. As a result, the electroscope leaves become further apart.

Thus, image B is representing the given scenario.

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Answer:kinetic energy

Explanation:

Answer:

P= phg

so:- the height is 10m

density of water 1000 kg/m3

gravity is 9.8m/s2

P=1000 kg/m3*10m*9.8m/s2

=98000Pa

=98KPa

The work done is 8427 J while the energy expended is  80.25 W.

Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;

W = mgh

Total mass of the water bucket and chain = 13.9 kg +  19.3 kg = 33.2Kg

Distance covered =  25.9 m

W = 33.2Kg * 9.8 m/s^2 * 25.9 m

W = 8427 J

Recall that the work done = Energy expended

Power = Energy expended/ Time

Power = 8427 J/ 1.75 * 60 seconds

Power = 80.25 W

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The minimum coefficient of static friction is  0.807

Static friction is friction between two or more solid objects that are not moving relative to each other.

According to the question,

When an object turns on a curve path at a specific speed, two forces will be acting on the object which balance each other:

Centripetal force and friction force.

The friction coefficient between the object and the path is independent of the object's mass.

Given,

The radius of a corner is, r=4.16m

The turning speed of the car is, v

[tex]20km/h * (\frac{\frac{5}{18}m/s }{1km/h} )[/tex]

= 5.72 m/s

To avoid slipping a coffee cup on the car's dashboard, the centripetal force would balance the friction force.

So,  Fc=Ff

[tex]\frac{mv^2}{r}[/tex]=μmg

μ=[tex]\frac{v^2}{rg}[/tex]

Here,

Fc represents the centripetal force.

Ff represents the friction force.

μ represents the coefficient of static friction.

g represents the gravitational acceleration whose value is 9.8m/s²

By Substituting the values in the above formula, we get:

μ =[tex]\frac{(5.72m/s)^2}{(4m)(9.8m/s^2)} \\[/tex]  ≈ 0.807

Therefore,

The minimum coefficient of static friction is  0.807

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

W = mg

W = 0.15 x 9.8 = 1.47 N

T = mg = 1.47 N

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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Answer:

The value of work being done on the object is 958J.

Explanation:

Work done is equal to force multiply by distance, but when the angle is between the force and distance work done=Force (cos theta) × distance

Work done = Force(cos theta) × distance

Work done = 25N(cos 40.0°) × 50m

Work done = 25N(0.7660) × 50m

Work done = 19.15N × 50m

Work done = 957.5J = 958J

Therefore the value of work being done on the object is 958J.

The x-component for the vector given in the image is : [tex]B_{x}[/tex] = 18.3m

A Vector can be defined as a quantity that possesses both magnitude and direction thereby giving more details.

A vector is very important when it comes to dealing with forces.

Analysis

[tex]B_{x}[/tex] = B cos(Θ)

Θ = 180° - 170° = 10°

[tex]B_{x}[/tex] = B cos(10°) = 18.3m

In conclusion, The x-component for this vector is : [tex]B_{x}[/tex] = 18.3m

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Answer:

True

Explanation:

it is true because blabla bla

The sun is the mother star of the solar system, which only emits light to half of the planet, while the other part is always dark.

The sun emits light towards the earth, which dominates all life on earth. The movements of the Moon around the Earth and of the Earth around the Sun are complex. Movements of rotation around their own axes are superimposed on movements of orbital translation. The Earth and the Moon rotate around their own axes: This is rotation.

Answer:

Power = 2000J/s

Explanation:

Power= Total Work done/Time

Given

mass = 100kg

distance(height) = 10m

time = 5s

acceleration due to gravity = 10m/s²

Work done = force × distance

Finding Force

Force = mass × acceleration due to gravity

Force = 100kg × 10m/s²

Force = 1,000N

Finding work done

Work done = Force × distance

Work done = 1,000N × 10m

Work done = 10,000J

Finding Power

Power = work done/time

Power = 10,000J/5s

Power = 2,000J/s

Therefore Power is 2,000J/s

There should be a reinstatement of the National Maximum Speed Limit for the safety of road users because the minimum braking distance is 2,25m.

Hence, we can see that the minimum braking distance for the car would be:

d'/d= (v'1/v1)^2 = (1.5/1)^2

= 2.25m

The speed limit ensures that drivers do not exceed a set speed limit in order to reduce road accidents and keep road users safe and also the pedestrians would be able to navigate easier.

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Answer:

Sparks often indicate the presence of a high voltage, or "potential field". The higher the voltage; the farther a spark can jump across a gap, and with enough energy supplied can lead to greater discharges such as a glow or an arc.

Explanation:

Answer:

The centripetal acceleration (ac)=314m/s²

Explanation:

look at the attachment ☝️

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

[tex]\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg[/tex]

We can simplify and rearrange the equation to solve for 'v'.

[tex]\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}[/tex]

Plugging in values:

[tex]v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}[/tex]

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

[tex]\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg[/tex]

Rearranging for 'T":
[tex]T = \frac{mv^2}{r} + mg\\\\[/tex]

Plugging in the appropriate values:
[tex]T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}[/tex]

a) mg[tex]\frac{1}{2}[/tex] = [tex]ml^\frac{2}{3}[/tex] * ω

b)ω = [tex]\frac{3g}{2l}[/tex]

c)14.7  the magnitude of the initial acceleration of the end B when the string is cut.

Acceleration is the rate of change of the velocity of an object with respect to time.

Given

A rod of length 56.0 cm and mass 1.40 kg is suspended by two strings which are 42.0 cm long, one at each end of the rod.

When the string is cut rod rotates about other end.

Let's take momentum equation about that end:

mg*[tex]\frac{1}{2}[/tex] = Iω [I = [tex]ml^\frac{2}{3}[/tex]]

a) mg[tex]\frac{1}{2}[/tex] = [tex]ml^\frac{2}{3}[/tex] * ω

Angular acceleration:

b)ω = [tex]\frac{3g}{2l}[/tex]

c)Acceleration of end B:

a = lω= [tex]\frac{3g}{2}[/tex] =[tex]\frac{3*9.8}{2}[/tex]

a = 14.7 [g=9.8]

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An object must move in the same direction as the force is applied for work to be accomplished. Option C is correct.

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative. it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.

The work done is found as;

W= F × d

Where.

F is the force

d is the displacement

W is the work done

In order for work to be done, an object must move in the same direction, and the force is applied

Hence option C is correct.

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The total work done is  5980 Joules and the power expended is 57 Watts.

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m  

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules  

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

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The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

The average power produced by the rocket is calculated as follows;

P = FV

where;

Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

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The formula for the impulse exerted on the ball at any given instant is the change in the momentum of the ball.

The impulse exerted on the ball at any given instant is calculated as follows;

J = ΔP

J = Pf - Pi

where;

Thus, the formula for the impulse exerted on the ball at any given instant is the change in the momentum of the ball.

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Answer:

Approximately [tex]4.61\times 10^{3}\; {\rm N}[/tex] upwards (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

External forces on this astronaut:

Let [tex](\text{normal force})[/tex] denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

[tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex].

Let [tex]m[/tex] denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be [tex](\text{weight}) = m\, g[/tex].

Let [tex]a[/tex] denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be [tex](\text{net force}) = m\, a[/tex].

Rearrange [tex]\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}[/tex] to obtain an expression for the magnitude of the normal force on this astronaut:

[tex]\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}[/tex].

Answer:

antibiotics

Explanation:

im not sure if this is right hope this helps lol

Answer:

[tex]\huge\boxed{\sf a = 0.7 \ m/s^2}[/tex]

Explanation:

Initial velocity = [tex]V_i[/tex] = 0.5 m/s

Final velocity = [tex]V_f[/tex] = 4 m/s

Time = t = 5 s

Acceleration = a = ?

[tex]\displaystyle a =\frac{V_f-V_i}{t}[/tex]

Put the givens in the formula.

[tex]\displaystyle a=\frac{4-0.5}{5} \\\\a = \frac{3.5}{5} \\\\a = 0.7 \ m/s^2\\\\\rule[225]{225}{2}[/tex]  

Assuming constant acceleration, the requisite magnitude is [tex]a[/tex] such that

[tex]\left(50\dfrac{\rm m}{\rm s}\right)^2 - 0^2 = 2a (2000\,\mathrm m)[/tex]

Solve for [tex]a[/tex] :

[tex]a = \dfrac{\left(50\frac{\rm m}{\rm s}\right)^2}{4000\,\mathrm m} = \boxed{0.625 \dfrac{\rm m}{\mathrm s^2}}[/tex]

The theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.

The theory of uniform accelerated motion is a scientific statement regarding the acceleration of an object in steady conditions.

This theory (uniform accelerated motion) is fundamentally applied in physic and astrophysics.

In conclusion, the theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.

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