An unbalanced force acting on an object will cause it to? A. Increase in mass. B. Decrease in mass. C. Accelerate. D. Remain at rest.

Answer:

See below.

Explanation:

According to the question, we know that,

work done is given by,  [tex]W=qV[/tex]

and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]

therefore equating both the equations we get,

[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]

m= mass of electron =  [tex]9.1*10^{-31} kg[/tex]

q= charge on an electron = [tex]1.6*10^{-19} C[/tex]

v= speed of electron= 700000m/s

substituting the values in the above equation, we get

[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]

(1).  the potential difference that stopped the electron is 1.39 volts.

now the kinetic energy equation is :  2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]

or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]

(2).  the initial kinetic energy of the electron is 1.39eV.

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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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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[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]

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:

The answer should be choice B.

The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.

A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.

This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.

The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.

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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]

The frequency of the 5 waves with a length of 4m hit the shore every 2 seconds is 2.5 Hz.

This is the number of cycles completed by a wave in one second. The s.i

unit of frequency is Hert (Hz).

From the question, to calculate the frequency of 5 waves with length of 4 m that hit the shores every 2 seconds, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The frequency of the wave is 2.5 Hz.

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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 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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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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Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

Answer: what I know that the unit is farad

Explanation:

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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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.

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

The athlete can improve performance by building strength, coordination and balance.

This is an individual who is proficient in sports and other forms of physical exercise.

Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.

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Answer:

Explanation:

Speed of light v = 3 x 10⁸ m/s

wavelength λ = 8.1 x 10⁻⁸ m

frquency f = v/λ = 3.7 x 10¹⁵ Hz

If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

C = λν

As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,

The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸

                                            =3.7 × 10¹⁵ Hz

Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz

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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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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:

5.98 kg

Explanation:

To solve this problem, let use the Linear Momentum Conservation Law:

Before collision: [tex]\sum p=p_{1}+p_{2}=m_{1}v_{1}+m_{2}v_{2}=(2.99v_{1})+0=2.99v_{1}[/tex]

After collision: [tex]\sum p'=p_{1}'+p_{2}'=(2.99+m_{2})(v_{1}/3)[/tex]

So, we obtain:

[tex]\sum p =\sum p' \rightarrow 2.99v_{1}=(2.99+m_{2})(v_{1}/3) \rightarrow 8.97 = 2.99 + m_{2}[/tex]

[tex]m_{2}=8.97-2.99=5.98 kg[/tex]

Answer: a = 0.647 m/s^2

Explanation:

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

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:

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.

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".

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:

0.5 mega grams

Explanation: