If I ride my bike at 10 mph and traveled 5 miles, how long did I ride in both hours and
minutes?

To better understand crash dynamics we have to look at "the laws of motion."

The laws of motion

The laws of motion were introduced by Sir Isaac Newton in 1687 in his book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), which defined the laws of motion, or three fundamental laws that govern the movement of bodies. The laws of motion, according to Newton, govern the motion of an object or a system of objects that interact.

It defines the concepts of force and mass, and the fundamental dynamics of motion.The following are the laws of motion:Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. The velocity of an object changes proportional to the force applied to it, and the acceleration of an object is proportional to both its force and its mass. For every action, there is an equal and opposite reaction.

Therefore, these laws are necessary to fully grasp crash dynamics because they explain how objects respond to outside forces that cause them to accelerate or decelerate.

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The resultant force when a 7 N force acts in the same direction as the 6 N force is 13 N.

When an object is subject to several forces, the resultant force is the force that act alone and produces the same acceleration as all those forces.

When two or more forces act in the same direction, their resultant force is the sum of all the forces.

To calculate the resultant force, we use the formula below,

Formula:

Where:

Substitute these values into equation 1

Hence the resultant force is 13 N.

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The pressure at the upper level of a water flow into the house by means of pipe is 1081 kPa.

Calculate the cross-sectional area of the lower pipe:

A₁ = πr₁²

where:

A₁ = cross-sectional area of the lower pipe (m²)

π = mathematical constant (3.14)

r₁ = radius of the lower pipe (m)

A₁ = π(0.12 m)² = 0.0452 m²

Calculate the cross-sectional area of the upper pipe:

A₂ = πr₂²

where:

A₂ = cross-sectional area of the upper pipe (m²)

π = mathematical constant (3.14)

r₂ = radius of the upper pipe (m)

A₂ = π(0.06 m)² = 0.0113 m²

Calculate the flow rate per unit area:

q = Q/A

where:

q = flow rate per unit area (m³/s)

Q = flow rate (m³/s)

A = cross-sectional area (m²)

q = 6 m³/s / 0.0452 m² = 13.28 m²/s

Calculate the velocity of the water in the lower pipe:

v₁ = q/A₁

where:

v₁ = velocity of the water in the lower pipe (m/s)

q = flow rate per unit area (m³/s)

A₁ = cross-sectional area of the lower pipe (m²)

v₁ = 13.28 m²/s / 0.0452 m² = 29.3 m/s

Calculate the velocity of the water in the upper pipe:

v₂ = q/A₂

where:

v₂ = velocity of the water in the upper pipe (m/s)

q = flow rate per unit area (m³/s)

A₂ = cross-sectional area of the upper pipe (m²)

v₂ = 13.28 m²/s / 0.0113 m² = 117.0 m/s

Calculate the head loss:

hL = (v₁² - v2₂²) / 2g

where:

hL = head loss (m)

v₁ = velocity of the water in the lower pipe (m/s)

v₂ = velocity of the water in the upper pipe (m/s)

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

hL = (29.3 m/s)² - (117.0 m/s)² / 2(9.8 m/s²) = 23.2 m

Calculate the pressure at the upper level:

p₂ = p₁ + ρghL

where:

p₂ = pressure at the upper level (Pa)

p₁ = pressure at the lower level (Pa)

ρ = density of water (1000 kg/m³)

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

hL = head loss (m)

p₂ = 400 kPa + 1000 kg/m³(9.8 m/s²)(23.2 m) = 1081 kPa

Therefore, the pressure at the upper level is 1081 kPa.

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Work done by the motor of London eye wheel is 5.67 x 10^4 J

For two reasons, dragging a load on wheels is much simpler than dragging it on the ground. Rotors lessen friction. The wheels dig in and rotate around strong rods called axles rather than just gliding over the surface.

I = mr2 = 1.9 x 10^6 x (67.5)2

            => 8.65 x 10^9

Work done => 1/2 x 8.65 x 10^9 x (3.62 x 10^-3)2

  => 5.67 x 10^4 J

Since the radian is a dimensionless quantity and the SI unit of angular velocity is radians per second, the SI units of angular velocity can be written as s1. The Greek letter omega (or, occasionally, ) is used to denote angular velocity.

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Answer:

Explanation:

A chemical change.

Usually those are irreversible. Or they may be reversible, but the form they take may leave your object not the same as they started out.

A physical change might be just as deadly. If the object melted like a chocolate Easter Bunny then the object would be irreversible as well. Take a better example.

Suppose you are talking about a Gold Coin. If you heated it so it melted, the gold would retain its value, but the fact that it is a coin and valuable as such, means that it has lost that part of its value.

I really don't know. My instincts tell me that the chemical change is more dangerous, but I can't rule out the other choice..

Hooke's Law can be given as follows sometimes:

The restoring force of a spring is equal to the spring constant multiplied by the displacement from its normal position:

F = -kx

Where, F = Restoring force of a spring (Newtons, N)

k = Spring constant (N/m)

x = Displacement of the spring (m)

The negative sign relates to the direction of the applied force and by convention, the minus or negative sign is present in F = -kx. The restoring force F is directly proportional to the displacement (x), according to Hooke's law. When the spring is compressed, the displacement (x) is negative. It is zero when the spring is at its original length and positive when the spring is extended.

Practically, Hooke's Law is applicable only within a limited frame of reference, and through experimenting, this statement proves to be true. Because materials cannot be compressed beyond a certain size or expanded beyond a certain size without some permanent deformation or change of their original state.

The law only applies under some conditions such as a limited amount of force or deformation. Factually, many materials will noticeably deviate from Hooke's law even before those elastic limits are reached.

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θ = 39.7º is the angle in degrees that the string makes with respect to the vertical using Newtons law.

In each exercise, we construct the equations at the equilibrium point using Newton's second law for the sphere. We'll assume that plate 1 is on the left for this exercise.

Y Axis

    Y= -W = 0 = W

X axis

       X= - F_{e2} + Tₓ = 0

Let's utilize trigonometry to determine the tension's component parts. We gauge the angle in relation to the vertical

        sin θ = Tₓ / T

        cos θ = T_{y} / t

        Tₓ = T sin θ

        T_{y} = T cos θ

Gauss's law can be used to determine the electric field of each leaf. Since a cylinder forms a Gaussian surface, the component of the field perpendicular to the cylinder's base is the one containing electric flow.

 F = ∫ E. dA

The flow is towards both sides of the plate in this instance, and the scalar product is reduced to the algebraic product.

F = 2E A = q_{int} / ε₀

let's use the concept of surface charge density

       σ = q_{int} / A

we substitute

       2E A = σ A /ε₀

         E = σ / 2ε₀

    T cos θ = mg

         - q σ₁ / 2ε₀  - q σ₂ /2ε₀  + T sinθ = 0

we introduce t in the second equations

         - q /2 ε₀  (σ₁ + σ₂) + (mg / cos θ) sin θ = 0

         mg tan θ = q /2ε₀   (σ₁ + σ₂)

         θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)

data indicates the mass of 0.26 g = 0.26 10⁻³ kg

give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²

let's calculate

        θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2  (6.25 10⁻¹² *0.26 10⁻³ 9.8))

        θ = tan⁻¹ 8.3 10⁻¹)

        θ = 39.7º

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The statement that is true about the forces acting on the car is that the net force acting on the car from all directions is zero. Option A is correct.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

The complete question is

"A car is moving at a constant speed along a straight line. Which statement is true about the forces acting on the car?

A.

The net force acting on the car from all directions is zero.

B.

The net force acting on the car is greater than the car's weight.

C.

The net force acting on the car is in the direction of the car's motion.

D.

The net force acting on the car is in the opposite direction of the car's motion."

The net force operating on the automobile in all directions is zero,

The statement that is true about the forces acting on the car is that the net force acting on the car from all directions is zero. Option A is correct.

Hence option A is correct.

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Answer:

It is A. Light blue

_____________

Hope this helps!

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┈┃┏┗┛┓┃╭┫o i n k    ┃

┈╰┓▋▋┏╯╯╰━━━━╯

╭━┻╮╲┗━━━━╮╭╮┈

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Answer:

The distance of the bar D = 1153 km

Explanation:

The electric force is the one that takes place between electric charges.

The electric force with which two point charges are attracted or repelled at rest is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them.

Recall that:

Electrical force(F) = I*B*L

where;

I = the current,

B = the magnetic field strength,

L = the length of the bar

However;

From the second equation of motion,

F = Ma

Since; (F) = I*B*L

Then,

Ma = IBL,

where;

M is the mass;

a is the acceleration

Making the acceleration (a) the subject of the formula, we have

a = IBL/M

Similarly;

From the third equation of motion;

v^2= u^2+2as,

where v and u are the final velocity and the initial velocity respectively

Here u = 0

Also; let distance s = D

Then

v^2 = 2aD

where;

a = IBL/M

Making the distance D  the subject of the formula, we get:

D = v^2/2a = v2*M/(2IBL)

D = 11200² × 20/(2×2300×0.86×0.55)

D = 1153047.155 m

D = 1153 km

The difference between the manufacturer's claims and your findings is, the manufacturer is concerned about amount of fat in the meat while you are concerned about calories.

Your findings concluded that only fat can provide calories needed in the body, perhaps other classes of food or nutrients in the meat can provide calories as well.

In the manufacturer's claim there are about 3% fat in the lunch meat and doesn't necessary mean that all the calories needed is the 3% fat.

Thus, there are lots of difference between your findings and the manufacturer's claim. The manufacturer is concerned about amount of fat in the lunch meat while you are concerned about calories in the meat.

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Answer:

3 Coulombs

Explanation:

Q = Current x time

Q = 2.0 x 1.5

Q = 3 Coulombs

Hi there!

22.

We can calculate the speed by first calculating the period of the stopper.

1 period = time taken for one revolution

10 rev/10 sec = 1 rev/sec

Since it takes 1 second for the stopper to make a complete circle, its velocity can be found using the equation for circumference:

\(v = \frac{2\pi r}{t} = \frac{2\pi (1)}{1} = \boxed{2\pi \approx 6.28 m/s}\)

23.

The centripetal force can be solved for using the following:

\(F_c = \frac{mv^2}{r}\)

m = mass (kg)

v = velocity (m/s)

r = radius (m)

Fc = Centripetal force (N)

Plug in the givens:

\(F_c = \frac{0.028(2\pi)^2}{1} = \boxed{1.105 N}\)

The final velocity of the 25.8 g marble after the collision is 16.15 cm/s.

The velocity of the 25.8 g marble after the collision is calculated as follows;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

The final velocity of the 25.8 g marble after the collision is calculated as;

( 25.8 x 21 ) + ( 12.4 x 13.8 ) = ( 12.4 x 23.9 ) + ( 25.8v )

712.92 = 296.36 + 25.8v

25.8v = 416.56

v = 416.56 / 25.8

v = 16.15 cm/s

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Answer:

The higher the amplitude, the higher the energy.

Given

Mass of the meteor, m=25000.0 kg

Mass of the satellite, m'=9600.0 kg

The speed of the meteor, v=250.0 m/s

The speed of the satellite, v'=375.0 m/s

To find

The speed after the collision

Explanation

Let the combined speed after collision be u.

a) they were initially moving towards each other

Let us consider the right direction to be positive.

Considering the meteor to be travelling towards the right and the satellite towards left.

By conservation of momentum

The combined speed is 76.58 m/s

b) they are moving in same direction

Let both the meteor and satellite moves towards the right direction

The combined speed is 284.68 m/s

Conclusion

a. The combined speed is 76.58 m/s

b. The combined speed is 284.68 m/s

Explanation:

\(\underline{\boxed{\large{\bf{Option \; A!! }}}} \)

Here,

We have to find the equivalent resistance of the circuit.

Here, \(\rm { R_1} \) and \(\rm { R_2} \) are connected in series, so their combined resistance will be given by,

\(\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ \)

\(\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ \)

\(\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ \)

Now, the combined resistance of \(\rm { R_1} \) and \(\rm { R_2} \) is connected in parallel combination with \(\rm { R_3} \), so their combined resistance will be given by,

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ \)

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ \)

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ \)

\(\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ \)

Reciprocating both sides,

\(\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ \)

Now, the combined resistance of \(\rm { R_1} \), \(\rm { R_2} \) and \(\rm { R_3} \) is connected in series combination with \(\rm { R_4} \). So, equivalent resistance will be given by,

\(\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ \)

\(\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ \)

\(\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ \)

\(\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ \)

\(\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ \)

Henceforth, Option A is correct.

\(\underline{\boxed{\large{\bf{Option \; B!! }}}} \)

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

\( \longrightarrow \) V = IR

\( \longrightarrow \) 3 = I × 3.33

\( \longrightarrow \) 3 ÷ 3.33 = I

\( \longrightarrow \) 0.90 Ampere = I

Henceforth, Option B is correct.

\( \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ \)

According to the graph below, the statements about these cars must be true are :

options B and C are correct

Velocity is described as the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

There are some similarities  between velocity and speed and they include:  

Velocity (v) is known as a vector quantity that measures displacement (or change in position, Δs) over the change in time.

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The graph is attached below:

Given:

The mass of the ball is,

The initial velocity of the ball is,

The angle with the floor is,

The change in momentum is the impulse by the floor, so the impulse is,

The impulse is 80 kgm/s along with the final velocity.

The tension in the cable during this motion will be 3750 lbs , when a steel cable carries a 3000 lbs load which is descending with a speed of 12 ft/s.

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Answer:

Total resistance in parallel is Rp

So (1/Rp) = (1/12)+(1/16)+(1/20)

Rp = ?

impedance is 25

25 = Rp + Rs

put the value of Rp you'll get the answer.

Answer: 59.8 or 299/5​- 55 x 4.8=264

Explanation:

Or if you wish to multiply it then the answer is above with the addition version.

Hope this helped :)

The initial velocity of the car in m/s is 26.41 m/s

We can convert 95.0 km/hr to m/s as illustrated below:

1 km/h = 0.278 m/s

Therefore,

95 km/h = 95 × 0.278

95 km/h = 26.41 m/s

Thus, the initial velocity in m/s is 26.41 m/s

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Answer:

The focal length of the mirror is 10.4 cm.

Explanation:

So substitute the values of object distance and image distance in the mirror equation,

1/f = 1/(-18cm) + 1/(-25.0cm)

1/f = -25cm/(-18cm x -25cm) - 18cm/(-18cm x 25cm)

1/f = ( -25cm - 18cm)/(18cm x 25cm)

1/f = -43.0/450.0

f = -10.4651 cm.

The focal length of the mirror is approximately -10.4 cm.

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The speed of skier's when he reaches the top plateau is 5.26 m/s, and the speed of skier's when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.

We can use the conservation of mechanical energy to solve this problem. Initially, the skier has kinetic energy, and at the top of the slope, he will have both potential and kinetic energy.

Since friction is negligible, the only force acting on the skier is the force of gravity. The work done by this force will be equal to the change in the skier's potential energy as he climbs up the slope. Therefore;

mgh = (1/2)mv²

where m = 104 kg is mass of the skier, g = 9.8 m/s² is the acceleration due to gravity, h = 1.2 m is height of the slope, and v is the skier's velocity when he reaches the top.

Solving for v, we get;

v = √(2gh) = √(29.81.2) = 5.26 m/s

Therefore, the skier's speed when he reaches the top plateau is 5.26 m/s.

In this case, there is also a frictional force acting on the skier, which does negative work on the skier as he moves up the slope. The work done by the frictional force is equal to the force of friction multiplied by the distance traveled;

W = Fd = μmgd

where μ = F/N is the coefficient of kinetic friction, N is the normal force acting on the skier (equal to the skier's weight), and d is the distance traveled up the slope (equal to the height of the slope, 1.2 m).

The net work done on the skier will be equal to the change in his mechanical energy;

Wnet = ΔK + ΔU = (1/2)m\(V_{f}\)² - (1/2)m\(V_{i}\)² + mgh

where vi = 8.7 m/s is the skier's initial velocity, \(V_{f}\) is his final velocity at the top of the slope, and ΔK and ΔU are the changes in kinetic and potential energy, respectively.

Since the net work done on the skier is equal to the work done by the gravitational force minus the work done by the frictional force, we have;

Wnet = mgh - μmgd

Substituting the expressions for Wnet and mgh, we get:

(1/2)m\(V_{f}\)² - (1/2)m\(V_{i}\)² = μmgd

Solving for \(V_{f}\), we get:

\(V_{f}\) = √(\(V_{i}\)² + 2μgd) = √(8.7² + 20.729.8×1.2) = 4.23 m/s

Therefore, the skier's speed when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.

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With the use of below formula, at 879 °C,  velocity will be double the velocity at 15 °C.

The velocity of sound waves in air is proportional to the square root of Thermodynamic temperature. That is, V = K\(\sqrt{T}\)

Given that the temperature at which the velocity of sound in air is twice its velocity at 15°C, Let us make use of the formula;

(v2/v1) = √(T2 / T1)

Where

Recall that absolute temperature = °C + 273.

If v2 = 2 × v1 and temperature in degree Celsius = 15°C, then,

Temperature in Kelvin K = 15 + 273 = 288

Substitute all the parameters into the formula

(2 × v1)/v1 = √(T2/288)

2 = √ (T2 /288)

Square both sides

4 = (T2/288)

T2 = 4 × 288

T2 = 1152K

Temperature in degrees Celsius = 1152 - 273 = 879 °C.

Therefore, at 879 °C,  velocity will be double the velocity at 15 °C.

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Answer:

SI unit of density - Kilogram per cubic metre (Kg/m³)

CGS unit of density - gram per cubic centimetre (g/cm³)