*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupants of the car appear to weigh 15% more than their normal weight

Answer: Gravitational Forces

Explanation:

Answer:

5.25 J

Explanation:

W = PE = (f)(x)

PE = 35N*0.15m

PE = 5.25 N*m

1 N*m = 1 J

PE = 5.25 J

Answer:

2.8 A

Explanation:

We know that;

Vrms = Vo/√2

Where;

Io = peak value of voltage = 200V

Hence;

Irms = 0.707 × 200

Irms = 141.4 V

Irms = Vrms/R = 141.4/50 = 2.8 A

Answer:

65N

Explanation:

Answer:

65 N

Explanation:

i just did the assignment on edge 2021

Answer:

the blue one

Explanation:

Answer:

3 m³ will occupy more volume.

Explanation:

We need to tell which occupies more volume : 3m³ or 100 ft³.

To compare, we first make the units same.

We know that,

1 m = 3.28 feet

1 m³ = 35.31 foot³

3 m³ = 105.94 foot³

Now, we can compare 105.94 foot³ and 100 ft³. It can be clear that 3 m³ will occupies more volume.

Answer:

the free-fall acceleration on the moon is 1.68 m/s^2

Explanation:

recall the formula for the gravitational potential energy (under acceleration of gravity "g"):

PE = m * g * h

replacing with our values for the problem:

46 J = 91 * g * 0.3

solve for the "g" on the Moon:

g = 46 / (91 * 0.3)

g = 1.68  m/s^2

Answer:

[tex]\mathrm{a.\:}123\:\mathrm{kg\cdot m/s},\\\mathrm{b.\:}95\:\mathrm{kg\cdot m/s},\\\mathrm{c.\:}28\:\mathrm{kg\cdot m/s}[/tex]

Explanation:

The momentum of an object is given by:

[tex]p=mv[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.

a)

The mass of the child and bike together is [tex]19+5.6=24.6[/tex] kilograms. Since they're moving at a velocity of 5.0 m/s, their momentum is:

[tex]p=24.6\cdot 5=\fbox{$123\:\mathrm{kg\cdot m/s}$}[/tex].

b)

The mass of the child is given as 19 kg. Since the child is on the bike moving at 5.0 m/s, it's implied the child is as well. Therefore, the momentum of the child is:

[tex]p=19\cdot 5=\fbox{$95\:\mathrm{kg\cdot m/s}$}[/tex].

c) The mass of the bike is given as 5.6 kg and it is moving at 5.0 m/s. Therefore, the momentum of the bike is:

[tex]p=5.6\cdot 5=\fbox{$28\:\mathrm{kg\cdot m/s}$}[/tex]

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined  for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

[tex]v=\frac{\Delta x}{\Delta t}[/tex]

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

[tex]v=\frac{37.1}{5.4}[/tex]

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

The velocity of the car will be 6.5 m/s.The rate of change of displacement is defined as speed.

The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

t is the time for camera deliver= 1.3 frames per second

l is the  length = 5.3 meters

The instantaneous velocity is given as;

[tex]\rm v = \frac{\triangle x }{\triangle t } \\\\ \rm \triangle x = 7 \times 5.3 \\\\ \rm \triangle x = 37.1 m[/tex]

The time engaged is find as;

1.3 frames = 1 s

[tex]\rm \triangle t= 7 \ frames \\\\ \rm \triangle t=5.4 sec[/tex]

Hence the velocity of the car driving;

[tex]\rm v= \frac{37.1}{5.4} \\\\ \rm v= 6.87 m/sec[/tex]

Hence the velocity of the car will be  6.5 m/s.

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Explanation:

Sun— Nuclear energy is converted to heat and light energy

Corn absorbs sunlight in order to produce sugars and oxygen for plant growth.

Corn is fermented to produce ethanol, which is burned to power generators to give an electrical current The speaker takes the electrical current and produces sound.

Answer:

sun, corn, corn, speaker

Explanation:

i took the k-12 test! have an amazing day! XX <3

Answer:

0J

Explanation:

work done= force x distance in direction of force.

since distance=0, work done=0

Answer:

By teh way is isn't it question of law or science and the picture is of what electric light or not I have read it so I was asking isn't is question of science

Given that,

Mass of Mr. Brooks = 100 kg

A chair lift up to 800 m to the top of the mountain.

To find,

How fast will he be going when he get back to the bottom of the mountain.

Solution,

Let v is the speed when he get back to the bottom of the mountain. Using the conservation of energy to find it.

[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 800} \\\\v=125.21\ m/s[/tex]

So, the required speed is 125.21 m/s.

Answer:

50km i think but i am not very sure

Answer:

A. Momentum of the grizzly bear is 3241.56 Kgm/s

B. The human will run at 28.56 m/s

Explanation:

A. Determination of the momentum of the grizzly bear.

Mass of grizzly bear = 420 lb

Velocity of grizzly bear = 17 m/s

momentum of the grizzly bear =?

Next, we shall convert 420 lb to Kg. This can be obtained as follow:

1 lb = 0.454 kg

Therefore,

420 lb = 420 lb × 0.454 Kg / 1 lb

420 lb = 190.68 Kg

Thus, 420 lb is equivalent to 190.68 Kg

Finally, we shall determine the momentum of the grizzly bear as follow:

Mass of grizzly bear = 190.68 Kg

Velocity of grizzly bear = 17 m/s

momentum of the grizzly bear =?

Momentum = mass × velocity

Momentum = 190.68 × 17

Momentum = 3241.56 Kgm/s

Thus, the momentum of the grizzly bear is 3241.56 Kgm/s

B. Determination of how fast the human will run.

Mass of human = 250 lb

Momentum of human = momentum of the grizzly bear

Momentum of the grizzly bear is 3241.56 Kgm/s

Momentum of human = 3241.56 Kgm/s

Velocity of human =?

Next, we shall convert 250 lb to Kg. This can be obtained as follow:

1 lb = 0.454 kg

Therefore,

250 lb = 250 lb × 0.454 Kg / 1 lb

250 lb = 113.5 Kg

Thus, 250 lb is equivalent to 113.5 Kg

Finally, we shall determine how fast the human will run as follow:

Mass of human = 113.5 Kg

Momentum of human = 3241.56 Kgm/s

Velocity of human =?

Momentum = mass × velocity

3241.56 = 113.5 × velocity

Divide both side by 113.5

Velocity = 3241.56 / 113.5

Velocity = 28.56 m/s

Therefore, the human will run at 28.56 m/s to match the momentum of the grizzly bear

Answer:

2.is the most massive object in the system, and less massive objects orbit more massive objects

is considered as a single focus by the objects describing elliptical paths about the sun.

Answer:

5 x 10⁻⁷N

Explanation:

Given parameters:

Mass of object 1  = 100kg

Mass of object 2 = 300kg

Distance  = 2m

Unknown:

Force of gravitational attraction between the objects  = ?

Solution:

Newton's law of universal gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

From Newton's law of universal gravitation we derive an expression:

       Fg  =  [tex]\frac{Gm_{1} m_{2} }{r^{2} }[/tex]

G is the universal gravitation constant = 6.67 x 10⁻¹¹

m is the mass

r is the distance between the bodies

Now insert the parameters and solve;

    Fg  = 6.67 x 10⁻¹¹ x [tex]\frac{100 x 300}{2^{2} }[/tex]    = 5 x 10⁻⁷N

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO  generally: (left or down) would be considered negative in an equation. And the other person  (right or up) would be considered positive. So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

0 is the number of equilibrium.

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

thanks for coming to my TED talk. I hope they don't delete this answer.

Answer:

450

Explanation:

Given,

Mass= 100kg

Velocity= 3 m/s

Kinetic Energy= ?

Kinetic Energy= 1/2 mv^2

= 1/2× 100× 3^2

= 1/2× 900

= 450.

HOPE IT HELPED :)

Answer:

a = 7.6\ mph/s

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

[tex]v_f=v_o+at[/tex]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

[tex]\displaystyle a=\frac{60 \ mph-0}{7.9}[/tex]

a = 7.6\ mph/s

Answer:

The equivalent resistance is 2 ohms.

Explanation:

let the first resistance = R₁ = 10 ohm

let the second resistance = R₂ = 10 ohm

let the third resistance = R₃ = 10 ohm

let the fourth resistance = R₄ = 10 ohm

let the fifth resistance = R₃ = 10 ohm

The equivalent resistance is calculated as;

[tex]\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \\\\\frac{1}{R_T} = \frac{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)}{R_1R_2R_3R_4R_5} \\\\R_T = \frac{R_1R_2R_3R_4R_5}{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)} \\\\R_T = \frac{(10^5)}{(10^4)+(10^4)+(10^4)+(10^4)+(10^4)} \\\\R_T = \frac{10^5}{5(10^4)} \\\\R_T = \frac{10}{5} \\\\R_T = 2 \ ohms[/tex]

Therefore, the equivalent resistance is 2 ohms.

Answer:

This is an example of Inelastic colission

Explanation:

Step one:

given:

mass of moose m1 = 620 kg

mass of train m2= 10,000kg

Initial velocity of moose  u1= 0 m/s

Initial velocity of train v1 = 10m/s

combined velocity of the system is given as v

Applying the conservation of momentum equation we have

m1u1+ m2u1= (m1+m2)V

substitutting we have

620*0+10000*10= (620+10000)V

100000= 10620V

divide both sides by 10620

V = 100000/10620

V=9.41m/s

The velocity of the moose after impact is 9.41m/s

Answer: 25%

Explanation:

From the question, we are informed that the temperature of source and sink of cannot engine are 400K

and 300K respectively.

The efficiency will be calculated as:

= 1 - t/T

= 1 - 300/400

= 1 - 3/4

= 1/4

= 25%

Answer:

Mechanical advantage = 3

Explanation:

Given:

Weight = 300 N

Force load = 100 N

Find:

Mechanical advantage:

Computation:

Mechanical advantage = Weight/Force Load

Mechanical advantage = 300/100

Mechanical advantage = 3

Mechanical advantage is the ratio of output force to the input force.It is also used to find the efficiency of the system.The value of mechanical advantage is 3.

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of  forces in levers and pulley.

It is an effective way of amplify the force in simple machines like lever  The theoretical mechanical advantage is defined as ratio of the force responsible for he useful work in system  to the applied force.

The theoretical value may be grater than the actual value of mechanical advantage because in the theoretical mechanical advantage friction is assumed to be absent.

The given data in the problem is;

weight that lifted (output)=300

Effort force(input) =100

Mathematically it is given as;

[tex]\rm M A= \frac{F_O}{F_I} \\\\\rm M A= \frac{300}{100}\\\\\rm M A=3[/tex]

Hence the value of mechanical advantage is 3.

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Answer:

A) conductors

Explanation:

A conductor can be defined as any material or object that allows the free flow of current or electrons (charge) in one or more directions in an electrical circuit. Some examples of a conductor are metals, tungsten, copper, aluminum, iron, graphite, etc.

Basically, the main purpose of a conductor in physics is to provide a low-resistance path between electrical circuits or components. This low-resistance path is to ensure that the electrical components allows the free flow of electrons and thus, enabling charge transfer.

Hence, the electrons in conductors move about more freely than the electrons in insulators which is why this type of material can be used to create electric circuits because it would significantly provide a low-resistance path between the electric circuits.

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

[tex]v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}[/tex]......(I)

The orbital speed of satellite B,

[tex]v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}[/tex]......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}[/tex]

Put the value into the formula

[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}[/tex]

[tex]\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}[/tex]

Hence, The ratio of their orbital speeds are 5:4.

Answer:

Angular velocity of merry-go-round is πk - 1 at t= T

Explanation:

From the question it is given that

[tex]\theta(t) = \pi k(t+k_e-\frac{t}{k} )[/tex] ..........................(1)

since mathematically, angular velocity is defined as

[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] ........................(2)

on substituing the value of θ(t) from equation 1 in equation (2) we get

[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] = [tex]\frac{d\pi k (t + k_e - \frac{t}{k} )}{dt}[/tex]   ............................(3)

on differentiating equation (3) with respect to time we get

ω(t) = πk(1 -[tex]\frac{1}{k}[/tex]) = πk - 1 angular velocity of merry-go-round

Therefore, angular velocity of merry-go-round is πk - 1 at t= T

The angular velocity of the merry-go-round at t=T is πk - 1. The pace at which angular displacement changes is defined as angular velocity.

The rate of change of angular displacement is defined as angular velocity.

It is stated as follows:

ω = θ t

The equation for the angular displacement is given as;

[tex]\theta (t) = \pi k (t+k_e-\frac{t}{k} )[/tex]

The angular velocity is found by the differentiation of the angular displacement.

[tex]\rm \omega(t) = \frac{d(\theta)}{dt} \\\\ \rm \omega(t) = \frac{d (\pi k (t+k_e-\frac{t}{k} )}{dt} \\\\ \rm \omega(t) = \pi k-1[/tex]

Hence the angular velocity of the merry-go-round at t=T is πk - 1.

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