How can an athlete participating in a 40m sprint modify and improve their performance based on the kinematic variable of speed and acceleration?

Answer:

The answer should be choice B.

Answer:

D. remain at rest

Explanation:

the x- component of the vector is 36.86 m.

Vectors are quantities that have both magnitude and direcion

To calculate the x-component of the vector, we use the formula below.

Formula:

Where;

From the question,

Given:

Substitute these values into equation 1

Hence, the x- component of the vector is 36.86 m.

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Based in the relationship between static frictional force and coefficient of static friction, the force of static friction is 24 N.

Friction is a force that opposes the relative motion of an object moving over another at their surafce of contact.

Frictional force is constant for each type of material. This constant is known as coefficient of friction.

The coefficient of friction is given as follows:

From the data given:

Force of static friction = 0.60 × 40

Force of static friction = 24 N.

Therefore, the force of static friction is 24 N.

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Answer:

Find the change in momentum of the upper stage, that is:

∆p = m(vf - vi)

m being the mass of the upper stage

vf being the final velocity which was given

vi being the initial which was also given

find ∆p

then use ∆p in the same equation

∆p being the answer you got above

m being the mass of the lower stage (given)

vi being the initial velocity (given)

vf being the final velocity of the lower stage which you were asked to find

Explanation:

During a collision the change in momentum (∆p) for both objects is equal regardless of their speeds or masses before or after the equation

Answer:

The static frictional force is the force between the magnetic forces statically and contains only two objects during this physical reaction. It results in electricity and a slight shock.

Answer:

A. An unbalanced force.

Explanation:

There needs to be a net force in order for the nail to be driven into presumably the wall. Without the net force then the hammer and nail wouldn't move.

Answer:

acceleration of the gravity

Explanation:

The weight of an object on the moon is less because the acceleration of the gravity on the moon is less.

The amplitude of the wave on the given sinusoidal wave graph is 10 cm.

The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.

Amplitude of the wave is calculated as follows;

From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.

Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.

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Answer: what I know that the unit is farad

Explanation:

The magnitude of the average force exerted on the ball by the club is 429.27 N.

Force can be defined as the product of mass and acceleration

To calculate the force exerted on the ball by the club, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The magnitude of the average force exerted on the ball by the club is 429.27 N.

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Biomolecules are fundamental for life. They required both energy supply and building structures.

Biomolecules are organic molecules that fundamental for life and must be ingested from the regular diet.

Biomolecules include proteins (e.g., meal proteins), lipids (fats) and also carbohydrates (e.g., glucose).

Biomolecules required both energy supply (e.g., glucose) and building structures (e.g., amino acids in proteins).

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Answer: a = 0.647 m/s^2

Explanation:

Acceleration = change in speed / time → a = 22 / 34 → a = .647 m/s^2

Answer:

0.5 mega grams

Explanation:

Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

Try this solution, all the details are in the attachment. note, the answer is marked with orange colour. If it is possible, check the provided solution in other sources.

Answer: ≈0.81c.

Hi there!

We can begin by summing the forces acting on the car.

Along the axis of the incline, the forces of friction and gravity are working. The force of friction points towards the top of the ramp, while the force of gravity works towards the bottom.

We can use trigonometry to determine the force due to gravity along the ramp.

[tex]F_g = Mg sin\theta[/tex]

The force due to friction is equal to:
[tex]F_f = \mu N[/tex]

The normal force is the vertical component of the force due to gravity, so:
[tex]F_f = \mu Mgcos\theta[/tex]

Now, the combination of these two forces produces a component of the centripetal force. Drawing a diagram, we see that the true centripetal force is the HYPOTENUSE, while these forces sum up to its horizontal component along the ramp.

Therefore:
[tex]F_c sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]

The centripetal force is equivalent to:
[tex]F_c = \frac{Mv^2}{r}[/tex]

m = mass (kg)

v = velocity (m/s)
r = radius (m)

Rewrite:
[tex]\frac{Mv^2}{r}sin\theta = Mgsin\theta - \mu Mgcos\theta[/tex]

Cancel out 'M'.

[tex]\frac{v^2}{r}sin\theta = gsin\theta - \mu gcos\theta[/tex]

Rearrange to solve for 'r'.

[tex]r = \frac{v^2sin\theta}{gsin\theta - \mu gcos\theta}[/tex]

Plug in values and solve.
[tex]r = \frac{(27.1^2)sin(29.1)}{(9.8)sin(29.1) - 0.4(9.8)cos(29.1)} = \boxed{266.365 m}[/tex]

This question involves the concept of  Newton's law of gravitation.

The force of gravity between the two spaceships is "1355.78 N".

According to Newton's Law of Gravitation:

[tex]F=\frac{Gm_1m_2}{r^2}[/tex]

where,

Therefore,

[tex]F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(500\ kg)(498\ kg)}{(35\ m)^2}\\\\[/tex]

F = 1355.78 N = 1.356 KN

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Two space ships traveling next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.

It is given that the First spaceship's weight ([tex]m_{1}[/tex]) is 500 kg,

The second spaceship's weight ([tex]\rm m_{2}[/tex]) is 498 kg.

The distance between spaceships (r) is 35 meters.

It is required to find the Force of gravity between these spaceships.

It is defined as the force which attracts any two masses in the universe.

By Newton's law of Gravitation:

[tex]\rm F= \frac{Gm_1m_2}{r^2}[/tex]  , Where

[tex]\rm F = The\ force \ of \ gravity \ between \ the \ spaceships\\\rm G= Universal\ Gravitational \ Constant = 6.67 \times 10^{-11} N.m^2/kg^2[/tex]

Putting values in the above formula:

[tex]\rm F = \frac{(6.67\times 10^{-11} N.m^2/kg^2)(500kg)(498kg)}{(35m)^2}[/tex]

F = 1355.78 N = 1.356  KN

Thus, Two spaceships travel next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.

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Answer:

m = 4.5 kg

Explanation:

w = - 1800 j

Fd = - 1800 j

mgd = - 1800 j

m = - 1800 ÷(gd)

m = - 1800 ÷( 10×-40)

m = 4.5 kg

Explanation:

conduction- putting a metal spoon in fire

convection- boiling water. In ventilation

Radiation- suns heat reaches us by radiation. heat from fire place reaches us by radiation.

What the equation given?

Maximum velocity occurs at the equilibrium position

So

Now

Now

As we know the formula

[tex]\\ \rm\rightarrowtail V_max=A\omega[/tex]

These expressions can be used

The maximum speed of the oscillating object will be given by [tex]V_{max}=Aw[/tex]

An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.

The given equation is as follows:

x(t)=Acost

Maximum velocity occurs at the equilibrium position x=0

x(0)=Acos0

x(0)=A

Hence the maximum velocity will be  [tex]V_{max}=Aw[/tex]

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[tex]\text{Given that,}\\ \\\text{Power,} ~P=24~ KW = 24000~ W\\\\\text{Force,} ~F=600~N\\\\\text{We know that,}\\\\P=Fv\\\\\implies v = \dfrac{P}{F} = \dfrac{24000}{600}=40~ ms^{-1}\\\\\\\text{It travels at 40 m/s}[/tex]

Answer:

A dot B = C     is the vector equation for this expression

A · B = A B cos θ

3 * 4 cos θ = 1       the value 1 is their dot product

cos θ = 1 / 12 = .083       θ = 85.2 deg

Answer:

If one wraps the fingers around the wire and points the thumb in the direction of the "conventional" current the fingers will point towards the North pole - the direction of the B-field.

In this case the B-field is pointed "West".

Hi there!

Part 1:

According to Newton's Third Law, every action has an EQUAL and OPPOSITE reaction.

For a collision, the objects involved exert an EQUAL and OPPOSITE impulse on each other, which means the forces exerted are equal as well.

Therefore, the school bus and Toyota Prius exert an EQUAL FORCE on each other.

Part 2:
The school bus has a greater mass.

Part 3:

Using Newton's Second Law:
[tex]\Sigma F = ma[/tex]

Using this relationship, the smaller the mass, the larger the acceleration. Since the forces are EQUAL, the Toyota Prius will experience a larger acceleration because it has a smaller mass.

The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.

To calculate the distance of the landing point of the grasshopper to the point where its jumps, we use the formula of range.

Range can be defined as the horizontal distance between the point of projection to the point where the projectile hit the plain again.

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The grasshopper is 1.47 m far away from the point where it jumps to the point where its lands.

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With the use of first and third equation of motion, the distance covered by a T-Rex is 30.51 m

When a body is in linear motion, the body is moving in a straight line. some of the parameters to consider are:

Given that a T-Rex move from 0 m/s to 9 m/s in 6.78 seconds, the distance covered can be found by calculating the acceleration.

Let us use equation 1

V = U + at

9 = 0 + 6.78a

a = 9 / 6.78

a = 1.33 m/[tex]s^{2}[/tex]

Now let us use equation 3

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as

[tex]9^{2}[/tex] = 2 x 1.33 x S

81 = 2.655S

S = 81/2.655

S = 30.51 m

Therefore, the distance covered by a T-Rex is 30.51 m.

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Answer:

Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.

Explanation:

Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].

Let [tex]m[/tex] denote the mass of the meteor.

Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.

Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.

Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.

The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:

Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.

During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.

Initial [tex]\text{KE}[/tex] of the meteor:

[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].

Initial [tex]\text{GPE}[/tex] of the meteor:

[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].

(Note the negative sign in front of the fraction.)

Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:

[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].

[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:

[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].

[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].

Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:

[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].

Substitute in the values and evaluate:

[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].

(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.