help me please I will thank you​

Answer:mechanical waves.

Explanation:

Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.

Answer:

8649 J

Explanation:

KE = 1/2mv^2

1/2(50)(18.6)^2

1/2(50)(345.96) = 8649 J

Answer:

The final velocity of the train and the moose after collision is approximately 19.51 m/s

Explanation:

The given mass of the moose, m₁ = 250 kg

The velocity of the moose, v₁ = 0

The mass of the oncoming train, m₂ = 10,000 kg

The velocity of the train, v₂ = 20 m/s

The velocity of the moose and the train after collision = v₃

By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Therefore, by substitution, we have;

250×0 + 10,000× 20 = (10,000 + 250) × v₃

200,000 = 10,250 × v₃

v₃ = 200,000/10,250 ≈ 19.51 m/s

The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s

Answer:

B

Explanation:

Hope this helps. :)

Answer:

320 degrees fahrenheit and 160 degrees celcius

Explanation:

there's a formula that i definately don't remember but it exists

Given values are:

Force,

Time,

Now,

→ [tex]Work \ done= f\times d[/tex]

                     [tex]= 100\times 50[/tex]

                     [tex]= 500[/tex]

hence,

The power will be:

= [tex]\frac{Work}{time}[/tex]

= [tex]\frac{500}{4}[/tex]

= [tex]125 \ watt[/tex]

Thus the response above is right.

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The acceleration at segment D is -20m/s²

The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)

Acceleration is the change in velocity with respect to time.

a = v-u/t

Acceleration at segment A:

Aa = 15-0/1-0

Aa = 15m/s²

Acceleration at segment B:

Ab = 20-15/2-1

Ab = 5m/s²

Acceleration at segment C:

Aa = 0-0/4-2

Aa = 0m/s²

Acceleration at segment D:

Ac = 0-20/5-4

Ac = -20m/s²

Hence the acceleration at segment D is -20m/s²

The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)

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Answer:

Carbon Dioxide.

Explanation:

Answer: the body is trained to wake up when the sun rises and get the energy from the sun and sleep and it is night on when the sun is not there. So when a watchman works at night there is no sun first of all to give him any kind of energy and secondly the body clock resists him waking up thus making him tired.

Explanation:

Answer:

I don't know the answer I hope you find it tho good luck##

Answer: 1.6 N

Explanation:

The weight of an object is calculated as:

W = g*m

where:

g = gravitational acceleration = 9.8m/s^2

m = mass of the object.

We also know that:

mass = density*volume.

or:

m = d*v

Let's start:

The weight of the beaker is 0.4N

And the weight of the beaker filled with water is 1.4N

Then the weight of the water alone will be:

1.4N - 0.4N = 1N = (d*v)*9.8m/s^2

And we know that the density of the water is:

1 g/cm^3

But we are working with Newtons, then we need to rewrite this with kilograms as the mass unit, we can use that:

1000g = 1kg

Now we can rewrite the density as:

d = 1 g/cm^3 = 1*(1/1000) kg/cm^3 = 0.001 kg/cm^3

Replacing that in the above equation, we get:

With this, we can find the volume that the water occupies.

W = 1 N = v* 0.001 kg/cm^3*9.8m/s^2

       1 N /( 0.001 kg/cm^3*9.8m/s^2 ) = 102.04 cm^3.

Now, when we fill it with a brine with a density of 1.2 g/cm^3, the mass of this brine in a volume of  102.04 cm^3 be:

M = (1.2 g/cm^3)*( 102.04 cm^3) = 122.448 g

Rewriting this in kg we get:

M = 122.448 g = (122.448/1000) kg = 0.122448 kg

Then the weight of this brine is:

M = 0.122448 kg*9.8m/s^2 = 1.2 N

And the beaker weighs 0.4N, then the beaker filled with this brine will weight:

1.2 N + 0.4N = 1.6 N

Answer:

as much as the wind is blowing has

Answer:

Him

Explanation:

He will push less but more and he will use more energy

Answer:

9.8×104Nm−2,0.025,3.92×106Nm−2

Solution :

Here, L=10cm=10×10−2m

F=100kgf=100×9.8N

ΔL=0.25cm=0.25×10−2m,

Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025

=3.92×106Nm−2

Explanation:

Answer:

By force and sloppy surface.

Explanation:

By applying force on the sled and slope of the path are the two ways the sled accelerate as it descends. If there is more friction between sled and the ground then force is required to push the sled to move downward while on the other hand, if the path on which sled moves is sloppy then it will move automatically without the use of force, so these two ways can accelerate the sled.

Answer:

You should face your fears sometimes, and get in front of a crowd do present your ideas. Not everyone is good at it, and many people get anxious when they have to speak in front of people but it’s best if you just do it and get it over with and maybe it won’t be as intimidating next time :)

Answer:

See the answers below.

Explanation:

The total power of the circuit is equal to the sum of the powers of each lamp.

[tex]P=60+100\\P=160 [W][/tex]

Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.

[tex]P=V*I[/tex]

where:

P = power [W]

V = voltage [V]

I = current [amp]

[tex]I = P/V\\I=160/240\\I=0.67 [amp][/tex]

So this is the answer for c) I = 0.67 [amp]

We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.

a)

[tex]V=I*R[/tex]

where:

V = voltage [V]

I = current [amp]

R = resistance [ohms]

Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.

[tex]P=V*I\\and\\V = I*R\\P = (I*R)*I\\P=I^{2}*R[/tex]

[tex]60 = (0.67)^{2}*R\\R_{60}=133.66[ohm] \\and\\100=(0.67)^{2} *R\\R_{100}=100/(0.66^{2} )\\R_{100}=225 [ohm][/tex]

b)

The effective resistance of a series circuit is equal to the sum of the resistors connected in series.

[tex]R = 133.66 + 225\\R = 358.67 [ohms][/tex]

Answer:

where is the picture. .......

Answer:

1.6m

Explanation:

To solve, we must use the equation: 1/f = 1/so + 1/si

f= focal point

so=object's distance

si=image's distance

We put in the numbers we have and we get: 1/f = 1/8.0 + 1/2.0.

1/f =  0.125 + 0.5 (changed the fractions to decimals)

1/f = 0.63 (0.625 rounded to the hundredths)

Now we solve for f:

1 = 0.63f

f = 1/0.63

f = 1.6

The focal length is 1.6m.

Hope this helped!

Answer:

Distance is the length of the path taken by an object whereas displacement is the simply the distance between where the object started and where it ended up.

Explanation:

Hope this helps! Good Luck on the rest of your assignments! :)

Answer:

Distance is quite literally the total amount of distance you have covered. Displacement is the total displacement from your origin. For example, if you were to move 5m north then 2m south, your total displacement from your origin would be 3m north.

The question is incomplete, the complete question is;

A car can go from 0 to 60 mph in 7.0 s. Assuming that it could maintain the same acceleration at higher speeds, how long would it take the car to go from 0 to 120 mph? A. 10 s B. 14 s C. 21 s D. 28 s

Answer:

B

Explanation:

We have;

vo = 0 m/s

v1 = 60 mph or 26.8 m/s

v2 = 120 mph or 53.6 m/s

t1 = 7s

To obtain a

v1 = vo + at1

but vo = 0 m/s

a = v1/t1 = 26.8 m/s/7s = 3.8 ms-2

To obtain t2 for the car;

v2 = vo + at2

but vo = 0 m/s

t2= v2/a = 53.6m/s/3.8ms-2

t2 = 14.1 s

Answer:

The answer is C) II and III only

Explanation:

Answer:

i read the whole thing the answer that matches the most is C

hope this helped :D

Explanation:

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

[tex]The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N[/tex]

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

[tex]W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N[/tex]

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

Answer:

21.67 m/s²

Explanation:

From the question given above, the following data were obtained:

Velocity (v) = 49.7 m/s

Radius (r) = 114 m

Centripetal acceleration (a) =?

The centripetal acceleration can be obtained by using the following formula:

a = v²/r

a = 49.7² / 114

a = 2470.09 / 114

a = 21.67 m/s²

Therefore, the centripetal acceleration of the car is 21.67 m/s²

Answer:

The answer is "False"

Explanation:

The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.

Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.

Answer:

161.28m

Explanation:

The resultant is calculated using the formula

R² = dx²+dy²

R = √dx²+dy²

From the diagram. dx = 101cos60°+85

dx = 101(0.5) + 65

dx = 50.5 + 85

dx = 135.5m

For the vertical component dy

dy = dsin theta

dy = 101 sin 60

dy = 101(0.8660)

dy = 87.47

R = √135.5²+87.47²

R = √18,360.25+7,651.0009

R = √26,011.2509

R = 161.28m

Hence the magnitude of the sum of the vectors is 161.28m

Answer:

The sum of the two vectors is 161.278 m.

Explanation:

Given;

vector, B = 101 m inclined at angle 60⁰

vector, A = 85 m inclined at angle 0⁰

Y-component of the vectors;

[tex]R_y = 101(sin \ 60^0) + 85(sin \ 0^0) =87.466 \ m[/tex]

X-component of the vectors;

[tex]R_x = 101(cos \ 60^0) + 85(cos \ 0^0)\\\\R_x = 50.5 + 85 = 135.5 \ m[/tex]

Sum of the two vectors;

[tex]R = A+ B\\\\R = \sqrt{R_y^2 + R_x^2} \\\\R = \sqrt{87.466^2 \ + \ 135.5^2} \\\\R = \sqrt{26010.55} \\\\R = 161.278 \ m[/tex]

Therefore, the sum of the two vectors is 161.278 m.

Answer:

Explanation:

The mass of the man can be found by using the formula

[tex]m = \frac{p}{v} \\ [/tex]

v is the velocity

p is the momentum

From the question we have

[tex]m = \frac{225.3}{2.65} \\ = 85.01886...[/tex]

We have the final answer as

Hope this helps you

Answer:

Equation for general relativity

Explanation:

Answer: 3.045 km/s

Explanation:

When an object is doing a circular motion, the velocity of the object is written as;

v = r*w

where;

r = radius of the circle

w = angular velocity.

In this case, we know that:

w = 7.25*10^(-5) s^-1

And the radius will be equal to the radius of the Earth, plus the height of the satellite, this is:

R = radius of the Earth + 35600 km = 6400km + 35600 km  = 42000 km

Then the velocity of the satellite will be:

v =  42000 km*7.25*10^(-5) s^-1 = 3.045 km/s

The velocity of a satellite with an orbit 35600 km above the surface of the earth and an angular velocity of 7.25×10–5 rad/sec = 3045 m/s

Velocity: This is the rate of change of displacement.

To solve this problem we need to use the formula for calculating the velocity of  an object in circular motion

The Velocity of the satellite is given as

V = ωr................ Equation 1

where V = velocity of the satellite, ω = angular velocity of the satellite, r = radius of the circle.

Note: r = height of the satellite above the surface of the earth+Radius of the earth

From the question,

Given: ω = 7.25×10⁻⁵ rad/sec, r = 35600+6400 = 42000 km = 4.2×10⁷ m

Substitute these values into equation 1

V = (7.25×10⁻⁵)(4.2×10⁷ )

V = 30.45×10²

V = 3045 m/s

Hence, the velocity of the satellite is 3045 m/s

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