A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car

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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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:

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:

Explanation:

while it is still falling it is turned into kinetic energy. While on the floor it turns onto potential energy. Remember energy cannot be created or the destroyed so the other ones are automatically out.

Answer:

0J

Explanation:

work done= force x distance in direction of force.

since distance=0, work done=0

Answer:

In picture one because string telephones are best heard when there is more tension on the string

Explanation:

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.

Learn more about the mechanical advantage refer to the link;

https://brainly.com/question/7638820?referrer=searchResults

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:

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 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:

the blue one

Explanation:

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:

gap junctions,tight junctions,and desmosomes

Answer:

[tex]v_2 = 6.406 / cos25.29[/tex]

Explanation:

From the question we are told that

Speed of snow ball 7.6m/s

Speed of chunk 9.3m/s at [tex]19 \textdegree[/tex]

Generally the equation for the conservation  of momentum is mathematically given as

Let the snow ball be b

and the chunk b/2

According to conservation of momentum we have

        [tex]b_u = (b /2) v_1 cos\theta_1 + (b / 2) v_2cos\theta_2[/tex]

        [tex]v_2 cos\theta_2 = 2 * 7.6 -9.3 * cos_19[/tex]

        [tex]v_2 cos\theta_2 = 15.2 -8.793 = 6.406[/tex]    

Therefore

          [tex](b/2)v_1 sin\theta_1 = (b/2) v_2sin\theta_2[/tex]

          [tex]v_2 sin\theta_2 = 9.3 * sin_19[/tex]

          [tex]v_2 sin\theta_2 = 3.027[/tex]

Mathematically From  above equations

        [tex]tan\theta_2 = 0.4726[/tex]

        [tex]\theta_2 = 25.29 \textdegree[/tex]

Therefore the speed and direction of second chunk

        [tex]v_2 = 6.406 / cos25.29[/tex]

        [tex]v2 = 7.0855m/s[/tex]

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:

F = Changing momentum / time

Explanation:

change in momentum = final momentum - initial momentum

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:

[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]

Answer:

  v_{f} = -0.693 m / s

Explanation:

The acceleration of the runner can be obtained from Newton's second law

       F = m a

        a = F / m

the bold are vectors, therefore the acceleration throughout the journey varies as the force has variations.

For the part of finding the velocity of the body we can use the relationship between the momentum and the variation of the momentum

            I = Δp

           ∫ F Δt = m [tex]v_{f}[/tex] - m v₀

           int F dt = m (v_{f}-v₀)

1) To find the change in velocity we must find the area under the curve of the graph, this can be done analytically if we know the functional of the curve or approximate it by intervals

a) between 0 <t <0.20 s

v) between 0.20 <t <0.30 s

a reasonable curve shape can be a Gaussian.

2) If we do not have the form of the cure, we can perform a graphical integration to find the area under the curve, we can do this by dividing the curve into small rectangles, finding the area of ​​each one and adding them.

3) Another even more approximate way is to create an average force in each interval and find the area of ​​this force, the average force is the average value of the force in the interval, let's use this method in the exercise

a) first interval 0 <t <0.20 Average force [tex]F_{mean}[/tex] = 300 N

    area = F_{mean} Δt

    area = I = 300 0.20

    I = 60 N s

the speed change is

    I = m Δv

    Δv = I / m

    Δv = 60/65

    Δv = 0.923 m / s

If we assume that the runner starts from rest, his final velocity is v = 0.923 m / s in the direction of the force.

b) second interval 0.2 <t <0.30s average force F_mean = 150 N

    area = I = 150 (0.30 - 0.20)

    I = 15 N s

the speed change is

    Δv = 15/65

    Δv = 0.23 m / s

Note that in this case the initial speed is not zero and since the two impulses are in the opposite direction the speed decreases

        [tex]v_{f}[/tex]= -0.923 + 0.23

        v_{f} = -0.693 m / s

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

Answer: Gravitational Forces

Explanation:

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:

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:

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:

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

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.

To learn more abouty the velocity refe to the link;

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Answer:

65N

Explanation:

Answer:

65 N

Explanation:

i just did the assignment on edge 2021