A 60.0 kg person jumps onto the floor from a height of 3.0m. Find
the KE of the jumper when she hits the ground.

Answer:

Light behaves differently in space than on Earth.

Explanation:

Because the gravity field is greater near earth than in most of space. Not the areas near stars, black holes, pulsars, and such but in the vast emptyness between the clumpy spots.

Answer:

The nervous system handles the stress response, which, if overworked, can eventually lead to diseases ranging from high blood pressure to diabetes.

Explanation:

hope I helped

9514 1404 393

Answer:

  1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

  2910d +308 = 2910×1.22

Solving for d, we find ...

  2910d = 2919(1.22) -308

  d = 1.22 -308/2910

  d ≈ 1.114 . . . kg/m³

The density of the hot air is about 1.114 kg/m³.

Answer:

The distance of man from the mirror is 3 m

Refer to the attachment

[tex]\Large\textsf{Hope \: It \: Helped}[/tex]

The distance of the man from the mirror is 3m

From the characteristics of image formed in a plane mirror,

⇒ From the last point,

⇒ From the question,

Hence, The distance of the man from the mirror is 3m

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Answer:

analyze, assess or interpret an historical event, era, or phenomenon,.

Explanation:

Secondary sources are works that analyze, assess or interpret an historical event, era, or phenomenon, generally utilizing primary sources to do so. Secondary sources often offer a review or a critique. Secondary sources can include books, journal articles, speeches, reviews, research reports, and more.

Explanation:

We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:

Vt = ωr with ω being the angular velocity.

We now calculate ω:

ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s

Now that we have ω we can calculate the centripetal aceleration:

a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2

The strength of the thrust is 122 newtons.

The motion of the rocket is described by the second Newton's law, whose model is shown below:

[tex]\Sigma F = F - D = m\cdot a[/tex] (1)

Where:

If we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:

[tex]F = D + m\cdot a[/tex]

[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 122\,N[/tex]

The strength of the thrust is 122 newtons.

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The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

W = 5 x 10

W = 50 N (downwards)

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N (upwards)

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

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The block will remain on the table because the normal force balances with the weight of the block. The correct answer is  50 N upward normal force

From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be

Weight W = mg

let g = 10 m/[tex]s^{2}[/tex]

W = 5 x 10

W = 50 N

The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,

Normal force N = 50 N

The block will remain on the table because the normal force balances with the weight of the block.    

Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.

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Answer:

The answer is "0.82 c".

Explanation:

Given:

Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]

Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]

The spacecraft velocity 2 measured by the Earth observation

   [tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]

            [tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]

Answer:

Explanation:

An impulse results in a change of momentum

If the wagon and dog both stop, they must have had equal and opposite momentums

FΔt = mΔv

F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)

v₁ = t₀ = 0

F = m(v₀)/t₁

F = 55(2.1)/0.1 = 1155 N

We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.

An object following a straight-line path at constant speed is option C.) has zero acceleration.

There are no forces operating on a body if it is travelling straight ahead at a steady speed. There are no forces operating on a body if it is travelling straight ahead at a steady speed.

Note that the physics concept of acceleration measures how quickly an object's motion is changing. An object's speed or velocity is what largely defines its motion.

Therefore, An object is considered to be accelerating when its velocity changes over time and as such  since acceleration of the object is  said to be  zero, one can say that the net force acting on it is also zero.

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Explanation:

decomposition reaction.....

Answer:

caco₃--cao+co₂

calcium oxide + carbon die oxide gives us calcium carbonate

this is the reaction of Acidic oxide(cao) and Basic oxide (co₂) to form salt.

Answer:

-9.0 × 10-6 m3 kg

Explanation:

I'm not sure if that's what you're looking for nor do I know how to explain it.

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

Answer:

h = change in vertical position (height)

has units of distance.

Explanation:

The equation below can be used to calculate a change in gravitational potential energy then the units used for the height would be meters.

The sum of all the energy in motion (total kinetic energy) and all the energy that is stored in the system (total potential energy) is known as mechanical energy.

As given in the problem, the equation below can be used to calculate a change in gravitational potential energy, the units used for the height of the object would be in meters, which is the SI unit of the length

The gravitational potential energy = mass×acceleration ×height

Thus, the unit of height used in the gravitational potential energy formula would be meter.

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Answer:

B

Explanation:

P1/P1 = 40/20

=2

Answer:

Looks good to me

Explanation:

#2 should probably be turning potential energy to kinetic.

Explanation:

a) Here is the free-body diagram. Note that I included the components of the weight mg (shown in dotted arrows) for use in the other parts of the problem.

b) The component of the weight parallel to the plane (shown in the diagram as a dotted arrow along the x-axis) is [tex]mg\sin15[/tex] and it is equal to

[tex]mg\sin15 = (25\:\text{kg})(9.8\:\text{m/s}^2)\sin15 = 63.4\:\text{N}[/tex]

c) Applying Newton's 2nd law to the y-axis, we can write

[tex]y:\;\;\;N - mg\cos15 = 0 \Rightarrow N = mg\cos15[/tex]

[tex]N = (25\:\text{kg})(9.8\:\text{m/s}^2)\cos15 = 236.7\:\text{N}[/tex]

d) The component of the weight mg into the plane is the same as the normal force, hence it's also 236.7 N.

e) To solve for the coefficient of friction, we apply Newton's 2nd law to the x-axis:

[tex]x:\;\;\;mg\sin15 - F_f = 0[/tex]

[tex]\Rightarrow F_f = mg\sin15\;\;(2)[/tex]

where [tex]F_f[/tex] is the frictional force defined as [tex]F_f = \mu N[/tex] so we can use Eqn(1) on Eqn (2) to write

[tex]\mu (mg\cos15) = mg\sin15[/tex]

Solving for [tex]\mu,[/tex] we get

[tex]\mu = \dfrac{\sin15}{\cos15} = \tan15 = 0.27[/tex]

Answer:

C. Mass

Explanation:

Answer:

  2.5 m/s east

Explanation:

Let east be the positive direction for velocity.

The change in momentum of the 0.75 kg model car is ...

  m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)

  = (0.75 kg)(20 m/s) = 15 kg·m/s

The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.

  m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s

The required relation is ...

  15 kg·m/s = -2(v2 -10) kg·m/s

  -7.5 = v2 -10 . . . . divide by -2

  2.5 = v2 . . . . . . . add 10

The velocity of the model truck after the collision is 2.5 m/s east.

Answer:

I THINK C

Explanation:

BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.

Answer:

Light takes less time than sound.

Explanation:

Let's say, the teacher and the student are at a distance "d" from each other.

The medium around them would be air.

And,

The speed of light in air is approx. 3× 10⁸ m/s

while, the speed of sound in air is approx. 330 m/s

We have a formula that establishes the relation between speed, distance and time.

[tex] \boxed{ \mathsf{speed = \frac{distance}{time} }}[/tex]

Our hunt for time — Speed in both the scenarios is known to us whereas the distance is same.

Sound

[tex] \mathsf{330 = \frac{d}{time_{s}} }[/tex]

[tex] \underline{\mathsf{time _{s} = \frac{d}{330} }}[/tex]

Light

[tex] \mathsf{3 \times {10}^{8} = \frac{d}{time _{l} } }[/tex]

[tex] \underline{ \mathsf{ time _{l} = \frac{d}{3 \times {10}^{8}} }}[/tex]

The best way of comparison is finding their ratio.

[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ \frac{d}{330} }{ \frac{d}{3 \times {10}^{8} } } }[/tex]

simplifying the fraction

[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{d \times (3 \times {10}^{8} )}{330 \times d}}[/tex]

d gets canceled and we're left with the following expression

[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ (3 \times10 \times {10}^{7} )}{330}}[/tex]

30, being a common factor in the numerator as well as denominator, gets canceled out. and in its place remains 1/ 11

(why?

=> 30÷330 = 1÷11)

[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ 1\times {10}^{7} }{11}}[/tex]

taking timeₛ to the numerator on the other side.

[tex] \implies \mathsf{time_{s} = \frac{ 1\times {10}^{7} }{11}\times time_{l}}[/tex]

Therefore, we get timeₛ is approx. 10⁶ times the timeₗ.

That's a big difference, no wonder light's way much faster than sound.

As lesser the time taken to cover a distance, faster is the wave.

The sound takes about 874,000 times MORE time than the light takes.

I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:

• its weight, mg, pulling it downward

• the normal force from contact with the road, n, pushing upward

• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)

By Newton's second law, the net forces on the car in either the vertical or horizontal directions are

∑ F (vertical) = n - mg = 0

∑ F (horizontal) = f = ma

where a is the car's centripetal acceleration, given by

a = v ²/r

and where v is the maximum speed you want to find and r = 25 m.

From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have

µmg = mv ²/r   ==>   v ² = µgr   ==>   v = √(µgr )

So the maximum speed at which the car can make the turn without sliding off the road is

v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s

Explanation:

12N by first law of newton is net force after colloision

Answer:

1.24

Explanation:

The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]

The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]

The wires have same resistance per unit length.

The resistance of a wire is given by :

[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]

According to given condition,

[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

The acceleration due to gravity on Mars is 11.81 m/s².

The given parameters:

The equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2[/tex]

where;

Since the time measured is two way time, the new equation for the total distance traveled is calculated as;

[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]

The acceleration due to gravity is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]

Thus, the acceleration due to gravity on Mars is 11.81 m/s².

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Answer:

2.quickly change the direction of your movement

Explanation:

(a) The period of the sound wave is 0.005 s.

(b) The wavelength of the wave when the speed of the wave is 1480 m/s is 6.73 m.

(c) The wavelength of the sound wave as it travels through air is 1.55 m.

The given parameters;

The period of the sound wave is calculated as follows;

[tex]T = \frac{1}{f} \\\\T = \frac{1}{220} \\\\T = 0.005 \ s[/tex]

The wavelength of the wave when the speed of the wave is 1480 m/s is calculated as follows;

[tex]v = f\lambda \\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{1480}{220} \\\\\lambda = 6.73 \ m[/tex]

The wavelength of the sound wave as it travels through air with a speed of about 341 m/s;

[tex]\lambda = \frac{v}{f} \\\\\lambda = \frac{341}{220} \\\\\lambda = 1.55 \ m[/tex]

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Answer:

I am serious about that

Explanation:

.......

Answer:  

Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].

Explanation:  

E inside [tex]= KQr_{1} /R^{3}[/tex]  

so if r1 will be the same then  

E  [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

Answer:

The electric field is E/8.

Explanation:

The electric field due to a solid sphere of uniform charge density inside it is given by

[tex]E =\frac{\rho r}{3}[/tex]

where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.

For case I:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]

So, electric field at a distance r is

[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]

Case II:

[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]

So, the electric field at a distance r is

[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]