What amount of force is required for a 200 kg grizzly bear to accelerate at 2.5 m/s/
s?

Answer:

Average speed = ( 2V + V1 + V2)/4

Explanation:

Given that a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.

Since the distance is covered at equal intervals of time, and

Speed = distance/time

For the first half distance,

V = distance/t

Cross multiply

Distance = Vt

For the second half distance

(V1 + V2)/2 = distance/t

Distance = t(V1 + V2)/2

The average speed = total distance/ total time.

Average speed = [Vt + t( V1 + V2)/2] ÷ 2t

Average speed = (2Vt + V1t + V2t)/4t

Average speed = t( 2V + V1 + V2)/4t

Time t will cancel out

Average speed = ( 2V + V1 + V2)/4

Answer:

The starting height of the ball is approximately 0.604 m

Explanation:

The given parameters are;

The mass of the the ball = 0.050

The speed with which it travels through the top loop = 2 m/s

The given height at which the ball moves at 2 m/s = 0.40 m

Therefore, we have;

1/2·m·v² = m·g·h

1/2·v² = g·h

h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204

The additional height = h = 0.204 m

Therefore;

The starting height of the ball  ≈ The given height at which the ball moves at 2 m/s + h

The starting height of the ball  ≈ 0.40 + 0.204 = 0.604 m

The starting height of the ball ≈ 0.604 m.

When any object in rest, then potential energy is present. But when object is in motion then object have kinetic energy.

Starting height of the ball is 0.604 m.

We know that, when any object is start from rest, then potential energy is converted into kinetic energy.

     [tex]\frac{1}{2}mv^{2} =mgh[/tex]

Where m is mass of object, g is gravitational acceleration , h is height and v is velocity of object. (value of g = 9.81 m/ second square)

from above equation,

we get,     extra height        [tex]h=\frac{v^{2} }{2g} \\\\h=\frac{4}{2*9.81}\\\\h=0.204[/tex] meter

The starting height of the ball will be sum of the height at which ball moves 2 m/s and extra height.

Starting height = 0.40 + 0.204 = 0.604 meter.

Learn more:

https://brainly.com/question/18963960

Answer:

Explanation:

I hope this helps

Answer:

White dwarf

Explanation:

Answer true

Explanation

Answer:

17.5 m/s

Explanation:

We can calculate the meters per second by dividing the distance by time. 3500 divided by 200 is 17.5, therefore the speed is 17.5 meters per second.

Answer:

All fluids exert pressure like the air inside a tire. The particles of fluids are constantly moving in all directions at random. As the particles move, they keep bumping into each other and into anything else in their path. These collisions cause pressure, and the pressure is exerted equally in all directions.

Explanation:

Answer:

b

Explanation:

Answer:

The answer is A) 29 meters

Explanation:

I got this question right on the test! :)

Answer:

Explanation:

From the information given:

We can properly determine the distance where the fish appear in the air viewing it from in front of the bowl by using the formula:

[tex]\dfrac{n_i}{d_o}+\dfrac{n_2}{d_1}= \dfrac{n_2-n_1}{r}[/tex]

where;

[tex]n_1[/tex] = refractive index in the air;  = 1.33  &

[tex]n_2[/tex] = refractive index in water. = 1

[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]

[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{4.7\ cm}[/tex]

[tex]\dfrac{1}{d_i}= - 0.26726 \ cm[/tex]

[tex]d_i =\dfrac{1}{ - 0.26726 \ cm}[/tex]

[tex]\mathbf{d_i }[/tex] = - 3.74 cm

2)

To determine where the fish appear to be when it is  38.9 cm from the front surface of the bowl by using the formula:

[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]

[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{38.9\ cm}[/tex]

[tex]\dfrac{1}{d_i}=- 0.0184759 \ cm[/tex]

[tex]d_i = \dfrac{1}{- 0.0184759 \ cm}[/tex]

[tex]\mathbf{d_i = }[/tex] -54.12  cm

Answer:

Wheres the charts??

Explanation:

Given :

A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10 s.

If the airplane is putting out an average force of [tex]5.8810\times 10^4 \ N[/tex].

To Find :

The average friction force exerted on the airplane by the air.

Solution :

Acceleration is given by :

[tex]a = \dfrac{162-120}{2.10}\ m/s^2\\\\a = 20 \ m/s^2[/tex]

Now, force equation is given by :

[tex]F - F_{friction} = ma\\\\F_{friction} = F-ma\\\\F_{friction} = 58810 - (2450\times 20 )\\\\F_{friction} = 9810\ N[/tex]

Therefore, frictional force exerted in the airplane by the air is 9810 N.

Answer:

A = 0.188 m²

Explanation:

First we find the distance between the plates by using the formula of electric field intensity:

E = ΔV/d

d = ΔV/E

where,

d = distance between plates = ?

ΔV = Potential Difference = 4 KV = 4000 V

E = Electric Field = 2 x 10⁸ V/m

Therefore,

d = 4000 V/(2 x 10⁸ V/m)

d = 2 x 10⁻⁵ m

Now, we find the Area of Plates by using formula of capacitance:

C = A∈₀∈r/d

where,

C = Capacitance = 0.25 μF = 0.25 x 10⁻⁶ F

A = Area of Plates = ?

∈₀ = Permittivity of free space = 8.85 x 10⁻¹² C/N.m²

∈r = Dielectric Constant = 3

Therefore,

0.25 x 10⁻⁶ F = A(8.85 x 10⁻¹² C/N.m²)(3)/(2 x 10⁻⁵ m)

A = (0.25 x 10⁻⁶ F)(2 x 10⁻⁵ m)/(3)(8.85 x 10⁻¹² C/N.m²)

A = 0.188 m²

Answer:

t = 0.0689 s

Explanation:

Given that,

Mass of a basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s (downward or negative)

Final velocity, v = 3.85 m/s (up of positive)

Average force, F = 72.9 N

We need to find the time it was in contact with the floor. The force is given by :

[tex]F=ma\\\\F=m\dfrac{v-u}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-(-4.23))}{72.9}\\\\t=0.0689\ s[/tex]

So, the time of contact is 0.0689 s.

Answer:Volume

Explanation:

Density = mass/ Volume

Answer:

Volume

Explanation:

Answer:

1.6 s

Explanation:

From the question given above, the following data were:

Velocity (v) = 15.68 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

Thus, we can calculate the time taken for Jack to hit the water by using the following formula:

v = gt

15.68 = 9.8 × t

Divide both side by 9.8

t = 15.68 / 9.8

t = 1.6 s

Therefore, it took Jack 1.6 s to hit the water.

Answer:

120km

Explanation:

Answer:

 v₂ = 2.5 v₁

car leaves at 2.5 times the speed of the minivan

Explanation:

This is an exercise of conservation of the momentum, to solve it we create a system formed by the minivan and the car, therefore during the crash the forces are internal and the momentum is conserved.

Initial instant. Before the crash

        p₀ = M v₁+ 0

Final moment. After the crash

        [tex]p_{f}[/tex] = M 0 + m v₂

how momentum is conserved

         p₀ = p_{f}

         M v₁ = m v₂

         v₂ = [tex]\frac{M}{m}[/tex] v₁

let's calculate

        v₂ = 2000/800 v₁

        v₂ = 2.5 v₁

therefore the car leaves at 2.5 times the speed of the minivan

Answer:

At the Berlin 2009 World Championships, Bolt set a world record time of 9.58 seconds for the 100m race, notching a top speed of 27.8 miles per hour (44.72 kilometers per hour) between meters 60 and 80, with an average speed of 23.5 mph.

Explanation:

Answer:

2m 13[tex]\frac{1}{3}[/tex]s

Explanation:

1.5m = 1s

200m = [tex]\frac{200}{1.5}[/tex] × 1s

          = 133[tex]\frac{1}{3}[/tex]s

          = 2m 13[tex]\frac{1}{3}[/tex]s

Answer:

μ = 0.125

Explanation:

To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.

Let's start working on the ramp

starting point. Highest point of the ramp

         Em₀ = U = m h y

final point. Lower part of the ramp, before entering the rough surface

        [tex]Em_{f}[/tex] = K = ½ m v²

as they indicate that there is no friction on the ramp

          Em₀ = Em_{f}

          m g y = ½ m v²

          v = [tex]\sqrt{2gy}[/tex]

we calculate

          v = √(2 9.8 0.25)

           v = 2.21 m / s

in the rough part we use the relationship between work and kinetic energy

          W = ΔK = K_{f} -K₀

as it stops the final kinetic energy is zero

          W = -K₀

The work is done by the friction force, which opposes the movement

          W = - fr x

friction force has the expression

          fr = μ N

let's write Newton's second law for the vertical axis

         N-W = 0

         N = W = m g

we substitute

            -μ m g x = - ½ m v²

           μ = [tex]\frac{v^{2} }{2 g x}[/tex]

Let's calculate

           μ = [tex]\frac{2.21^{2}}{2\ 9.8\ 2.0}[/tex]

           μ = 0.125

A 8.0 kg box is released from rest at a height y0 = 0.25 m on a frictionless ramp. The box slides from the ramp onto a rough horizontal surface. The box slides 2.0 m horizontally until it stops.

What is the friction coefficient of the horizontal surface?

Answer: 0.125

Answer:

A

Explanation:

Trust me I just took it !

heat food, warm water

Answer:

2.4J

Explanation:

Given parameters:

Spring constant  = 120N/m

Extension  = 0.2m

Unknown:

Amount of energy  = ?

Solution:

The energy stored in this stretched spring is called the elastic potential energy.

It can be derived using the expression below:

 Elastic Potential energy  = [tex]\frac{1}{2}[/tex] ke²  

k is the elastic constant

e is the extension

  Insert the parameters;

        Elastic potential energy  =  [tex]\frac{1}{2}[/tex]  x 120 x 0.2²  = 2.4J

Answer:

E =F.d =[1/2]mv^2

mad = [1/2]mv^2

d= v^2/2a ,v=u+at , v^2 = [at]^2 since u=0

So d = at^2/2

F = ma= 20a= 50 , a=5/2 and t=2

so d = [5/2][2^2]/2=5

Explanation:

Every action has an equal and opposite reaction. It is an action-reaction principle. Therefore the table exerts a force of 5 N on the book in order to be in stable condition.

Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.

Therefore the table exerts a force of 5 N on the book in order to be in stable condition.

The given data in the problem is ;

W is the weight of the book sits on table = 5N

N is the normal force on the book

From the equilibrium equation ;

Weight -Normal force on the book  =0

Weight =Normal force on the book

The normal force on the book =5N

Hence the table exerts a force of 5 N on the book in order to be in stable condition.

To learn more about Newton's third law of motion refer to the link;

https://brainly.com/question/1077877

Answer:

If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.

Explanation:

you can double the net force

Answer:

C

Explanation:

Answer:

The mission will help scientists investigate how planets formed and how life began, as well as improve our understanding of asteroids that could impact Earth.

Explanation:

Hope this helps :)

Answer:

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

Explanation:

Answer: i think its b

sorry if im wrong

Explanation:

Answer:

t = 9.14 s

Explanation:

We first analyze the accelerating motion by applying first equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car before turning off engine

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 25 m/s²

t₁ = time taken in accelerating motion

Therefore,

Vf₁ = 25t₁   ---------- equation (1)

Now, we apply second equation of motion:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = distance covered during accelerating motion

Therefore,

s₁ = (0)t₁ + (1/2)(25)t₁²

s₁ = 12.5 t₁²   ----------- equation (2)

Now, we analyze the decelerating motion by applying first equation of motion:

Vf₂ = Vi₂ + a₂t₂

where,

Vf₂ = Final Speed of Car = 0 m/s

Vi₂ = Initial Speed of Car after turning off engine

a₂ = deceleration of car = - 3 m/s²

t₂ = time taken in decelerating motion

Therefore,

Vi₂ = 3t₂   ---------- equation (3)

Now, we apply second equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered during decelerating motion

Therefore,

s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²

s₂ = Vi₂ t₂ - 1.5 t₂²  

using equation (3):

s₂ = 3 t₂² - 1.5 t₂²

s₂ = 1.5 t₂²   ------------ equation (4)

Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):

Vf₁ = Vi₂

using equation (1) and equation (3):

25 t₁ = 3 t₂

t₁ = 0.12 t₂   ------------ equation (5)

Also, we know that sum of the distances is 200 m:

s₁ + s₂ = 200

using equation (2) and equation (4):

12.5 t₁² + 1.5 t₂² = 200

using equation (5):

12.5 (0.12 t₂²) + 1.5 t₂² = 200

3 t₂² = 200

t₂² = 200/3

t₂ = 8.16 s

substitute this in equation (5):

t₁ = 0.12(8.16 s)

t₁ = 0.97 s

Hence, the minimum time required for this motion is:

t = t₁ + t₂ = 0.97 s + 8.16 s

t = 9.14 s