21. A toy car starts from rest and begins to
accelerate at 11.0 m/s2. What is the toy car's
final velocity after 6.0 seconds?

Answer:

Her speed at the bottom of the slide is 7.42 m/s

Explanation:

From the question,

The swimmer starts at rest, that is, her initial speed, u is 0 m/s.

Since she slides without friction and descends through a vertical height, then it is a free fall motion (due to gravity).

Also, from the question,

She descends through a vertical height of 2.81 m.

To determine her speed at the bottom of the slide, that is her final speed,

From one of the equations of motion for freely falling bodies

v² = u² + 2gh

Where v is the final speed

u is the initial speed

g is acceleration due to gravity (g = 9.8 m/s²)

and h is height

From the question,

u = 0 m/s

h = 2.81 m

Putting the values into the equation

v² = u² + 2gh

v² = 0² + 2×9.8×2.81

v² = 55.076

v =√55.076

v = 7.42 m/s

Hence, her speed at the bottom of the slide is 7.42 m/s.

Answer:

m = 50 [kg]

Explanation:

In order to solve this problem we must be clear about the difference between weight and mass. Weight is the product of mass by the acceleration of the planet or the star. While the mass is always preserved it never changes regardless of where it is located.

So for the earth we have:

g = gravity acceleration = 9.8 [m/s^2]

m = mass [kg]

W = weigth = 490 [N]

therefore the mass will be:

m = W/g

m = 490/9.8

m = 50 [kg]

Now it is important to remember that the mass will be the same on the moon or on the earth, but the weight will be different, because the gravity acceleration of the moon is different from the gravity acceleration on earth

So the gravity on the moon is equal to:

81.7 = 50 * gm

gm = 1.634 [m/s^2]

This question is incomplete

Complete Question

m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.

a) if the system is released from rest what will be its acceleration

Answer:

0.7 m/s²

Explanation:

The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.

(a) if the system is released from rest what will be its acceleration

g = acceleration due to gravity = 9.81 m/s²

Coefficient of Kinetic Friction = μk = 0.30

m1 = 10kg

m2 = 4.0kg

The formula to solve question a is given as:

a = acceleration at rest

m2g- μk m1g = (m1+ m2) a

Making a the subject of the formula:

a = (m2g- μk×m1g )/(m1+ m2)

a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)

a = 0.7 m/s²

v² - u² = 2 a ∆x

where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.

So

v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)

v² = 4645 m²/s²

v ≈ 68.15 m/s

Answer:

3

Explanation:

Answer:

  a) about 14.577 kg

  b) 300 N

Explanation:

b) In order for the acceleration to be the same for each mass, the 800 N force must be divided between the boxes in proportion to their mass. That is, the net force on the 5M mass must be 5/8 of the total force, or 500 N. Then the tension in the rope is 800 N -500 N = 300 N, which is 3/8 of 800 N.

Tension: 300 N

__

a) The total mass is 8M, and the total normal force on the floor is ...

  F = ma = (8M)(9.8 m/s^2)

The friction force is 0.7 times this, and is equal to the 800 N force pulling on the boxes.

  800 N = (8M)(9.8 m/s^2)(0.7)

  M = 800/(8·9.8·0.7) kg ≈ 14.577 kg

C.1cm

Explanation:

precision is how close two or more measurements are to each other

4?

Explanation:

sorry im not sure but

You can always take a metre ruler as a starting point. Your metre ruler has the same precision as your 15.0cm or 30.0cm ruler, so bring a ruler during exams as they'll come in handy ;)

The order goes like this:

rulers: 0.1cm or 1mm

measuring tape: 0.01cm or 0.1mm

vernier calipers: 0.001cm or 0.01mm

micrometer screw gauge: 0.0001cm or 0.001mm

((if i'm not wrong))

This question is incomplete, the complete question is;

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.

How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°

Answer:

the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J

Explanation:

Given that;

m = 9.9 kg

h = 4.9 m

d = 5 m

μ = 0.3

θ = 36.87°

Now from conservation of energy, the energy is;

Et = mgh

we substitute

Et = 9.9 × 9.8 × 4.9

= 475.398 J

Also the loss of energy i

E_loss = (umg cosθ) d

we substitute

E_loss  = 0.3 × 9.9 × 9.8 × cos36.87°  × 5

= 116.423 J

so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be

E = Et - E_loss

E = 475.398 J - 116.423 J

E = 358.975 J

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]acceleration = \frac{force}{mass} \\[/tex]

From the question

force = 20 N

mass = 4 kg

We have

[tex]a = \frac{20}{4} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

John Sebastian Bach

Answer:

Johann Sebastian Bach

Explanation:

please tell me if i am wrong! thank you!

Answer:

The car's average acceleration is [tex]10\ m/s^2[/tex].

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being vo the initial speed, a the constant acceleration, vf the final speed, and t the time, the following relation applies:

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

If we need to find the acceleration, we solve the above equation for a:

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

The car accelerates from rest (vo=0) to vf=70 m/s in t=7 seconds. Substitute the values into the formula:

[tex]\displaystyle a=\frac{70-0}{7}=\frac{70}{7}=10[/tex]

[tex]a=10\ m/s^2[/tex]

The car's average acceleration is [tex]10\ m/s^2[/tex].

Note: The choices are not very clear, but the second choice seems to be the correct answer.

Answer:

At constant velocity, his weight equals the force of friction. In other words, there is no net force. If however, he loosens his grip and decreases the friction force, he will accelerate downward.

Explanation:

Answer:

The horizontal displacement is 20 m.

Explanation:

Given that,

Velocity = 10 m/s

Height = 20 m

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Put the value into the formula

[tex]20=0+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]t^2=\dfrac{20\times2}{9.8}[/tex]

[tex]t=\sqrt{\dfrac{20\times2}{9.8}}[/tex]

[tex]t=2.0\ sec[/tex]

We need to calculate the horizontal displacement

Using formula of horizontal displacement

[tex]\Delta x=v_{x}\times t[/tex]

Put the value into the formula

[tex]\Delta x=10\times2.0[/tex]

[tex]\Delta x=20\ m[/tex]

Hence, The horizontal displacement is 20 m.

Answer:

kinetic and potential energy

Explanation:

Explanation:

Work done by a force is given by :

[tex]W=Fd\cos\theta[/tex]

Where

F is force, d is displacement and [tex]\theta[/tex] is the angle between F and d.

In this problem, a crane is moving in downward direction, the force gravity is in downward direction and the tension is in upward direction.

We know that if force and displacement is in same direction, work is positive while if force and displacement is in oposite direction, work is negative.

I would mean that, [tex]W_g[/tex] is positive, because gravity is parallel to the displacement and [tex]W_t[/tex] is negative, because the tension is opposite to the displacement.

electricity

Explanation:

the position (2,o

Answer:

N2 + 3H2 ----->  2NH3

Explanation:

Reactants side:

2 Nitrogen

5 Hydrogen

Products Side:

2 Nitrogen

5 Hydrogen

Answer:

50 m

Explanation:

The relationship between the intensity of sound in dB and distance is given by the formula:

[tex]B_2=B_1+20log(\frac{R_1}{R_2} )\\\\Where \ B_2\ is \ the\ sound\ intensity\ at\ distance\ R_2\ and\\B_1\ is \ the\ sound\ intensity\ at\ distance\ R_1\ \\\\Given\ that: B_1=100\ dB, R_1=5\ m, B_2=80\ dB\\\\B_2=B_1+20log(\frac{R_1}{R_2} )\\\\80=100+20log(\frac{5}{R_2} )\\\\-20=20log(\frac{5}{R_2} )\\\\log(\frac{5}{R_2} )=-1\\\\\frac{5}{R_2}=10^{-1}\\\\\frac{5}{R_2}=0.1\\\\R_2=5/0.1=50\ m[/tex]

Answer:

A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B)  T = 4.7 10⁴ K, C) n₂ = 42

Explanation:

A)  For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.

Let's reduce the units to the SI system

        E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J

The kinetic energy of the electron is

        K = ½ m v²

         E₀ = K

         v = √ 2E₀ / m

         v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)

         v = √ (2.14857 10¹²)

         v = 1.47 10⁶ m / s

now the speed of a calcium ion is asked, let's find sum

        m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg

        v = √ (2E₀ / M)

         v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)

        v = √ (0.2994457 10⁸)

        v = 0.5426 10⁴ m / s

B) the terminal energy of an ideal gas is

             E = 3/2 kT

              T = ⅔ E / k

              T = ⅔ (9,776 10-19 / 1,381 10-23)

              T = 4.7 10⁴ K

C) To calculate the energy of these lines we use the Planck expression

              E = h f

where wavelength and frequency are related

              c =λ f

              f = c /λ

let's substitute

              E = h c /λ

let's look for the energies

λ = 396.8 nm

  E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹

           E₁ = 5.0126 10⁻¹⁹ J

λ = 393.3 nm

           E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹

           E₂ = 5.0572 10⁻¹⁹ J

The difference in energy between these two states is

          ΔE = E₂ -E₁

          ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J

          ΔE = 0.0446 10⁻¹⁹ J

let's reduce eV

         ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

         ΔE = 2.787 10⁻² eV

Now let's use Bohr's atomic model for atoms with one electron,

               E = -13.606 Z² / n²

where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium

               n = √ (13.606 Z² / E)

λ = 396.8 nm

E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV

               n₁ = √ (13.606 20² / 3.132875)

               n₁ = 41.7

since n must be an integer we take

               n₁ = 42

λ = 393.3 nm

E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV

              n₂ = √ (13.606 20² / 3.16075)

              n₂ = 41.5

Again we take n as an integer

               n₂ = 42

We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin

Answer:

Total energy = Kinetic Energy + Potential Energy = Constant

Since the potential energy is lowest at point D the kinetic energy will be greatest at point D and the velocity will be the greatest.

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:

Given that,

[tex]m_{1}[/tex] = 18.5 kg

d = 2.3m

To find,

Acceleration of block 1:

∑[tex]F = ma[/tex]

⇒ -m₁gμ = m₁a

⇒ a = -gμ

⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]

∵ a [tex]= -2.16 m/s^2[/tex]

Now,

To determine the velocity of block one prior to the collision:

We know,

The initial velocity of block 1 = 8.25 m/s

⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx

⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]

∵ [tex]v = 7.63 m/s[/tex]

We also know,

[tex]m_{2}[/tex] = 7.25 kg

Now,

The velocity of block 2 post collision:

⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision

Through this,

⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]

∵[tex]v = 19.5 m/s[/tex]

The distance can be found through:

⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]

⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]

∵ Δ[tex]x = 87.8 m[/tex]

Thus, 19.5 m/s and 87.8 m are the correct answers.

Learn more about "Friction" here:

brainly.com/question/13357196

Answer:

Pressure increases as you move deeper below earth's surface.

Tempurature  increases as you move deeper below earth's surface.

Hope this helps!

Explanation:

Answer: f= M×A

1.75kg×24= 42N

Explanation:

Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!

Answer:

maibi.... D

Explanation:

I think is D

Answer:

The Energy transferred to the engine by burning gasoline = 216.67 KJ

Explanation:

The parameters given are:

The efficiency of the car engine, E = 30% = 0.3

Mass, m = 1300 kg

Initial velocity, u = 0, since the car is from rest

The final velocity, v = 10 m/s

Since the car was moving, we calculate its kinetic energy.

kinetic energy = ((1/2) (m) (v^2)

((1/2) (1300 kg) (10 m/s^2)

= 65,000 j

The Energy, Q transferred to the engine by burning gasoline in this case

= potential energy / The efficiency of the car engine, E

Q = 65,000 j / 0.3

= 216,666.66 J

Converting Joule to kilojoule

where 1KJ = 1000j

216,666.66 J = 216.67 KJ