A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

Answers

Answer 1
Answer:

Question:

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?

Statement:

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.

Solution:

  • Initial velocity (u) = 70 m/s
  • Acceleration (a) = -14 m/s^2
  • Time (t) = 3 s
  • Let the velocity of the car after 3 s be v m/s
  • By using the formula,

v = u + at, we have

v = 70 + ( - 14)(3) \n  =  > v = 70 - 42 \n  =  > v = 28

  • So, the velocity of the car after 3 s is 28 m/s.

Answer:

The car's speed after 3 s is 28 m/s.

Hope it helps


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Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

Answers

Answer:

The speed of the arrow immediately after it leaves the bow is 38.73 m/s

Explanation:

given information:

force,  F = 150 N

x = 50 cm = 0.5 m

mass of arrow, m = 50 g = 0.05 kg

We start from the force of the spring

F = kx

k = (F)/(x)

  = (150)/(0.5)

  = 300 N/m

The potential energy, EP of the spring is

EP = (1)/(2) kx^(2)

the kinetic energy, EK of the spring

EK = (1)/(2) mv^(2)

According to conservative energy,

EP = EK

(1)/(2) kx^(2) = (1)/(2) mv^(2)

kx^(2) = mv^(2)

v^(2) = (kx^(2) )/(m)

v = x\sqrt{(k)/(m) }

  = 0.5\sqrt{(300)/(0.05) }

  = 38.73 m/s

Final answer:

Using Hooke's Law, we can determine the speed of the arrow. The speed of the arrow immediately after it leaves the bow is approximately 38.7 m/s.

Explanation:

In this problem, we can use Hooke's Law to determine the speed of the arrow. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring:

F = -kx

Where F is the force, x is the displacement, and k is the spring constant.

In this case, the force exerted by the bow on the arrow is acting like a spring force. The force of the bow is 150N, and the displacement is 50cm (which is equivalent to 0.5m). So we can set up the equation as:

150N = -k * 0.5m

Now we can solve for k:

k = -150N / 0.5m = -300 N/m

Now that we have the spring constant, we can use it to find the potential energy stored in the bow:

PE = 0.5kx^2 = 0.5*(-300N/m)*(0.5m)^2 = 37.5 J

Next, we can use the conservation of energy to find the kinetic energy of the arrow right after it leaves the bow. The potential energy stored in the bow is converted into kinetic energy:

KE = PE = 37.5 J

The kinetic energy is given by the equation:

KE = 0.5mv^2

Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for v:

v = sqrt(2KE/m) = sqrt(2*37.5 J / 0.05 kg) = sqrt(1500) m/s ≈ 38.7 m/s.

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At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the tires rotating? Part (b) What is the centripetal acceleration at the edge of the tire in m/s^2?

Answers

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         \omega =(72)/(0.445)=161.8rad/s

Frequency

         f=(2\pi)/(\omega)=(2* \pi)/(161.8)=0.0388rev/s=2.33rev/min

Revolutions per minute = 2.33

b) Centripetal acceleration

               a=(v^2)/(r)

  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m

Substituting

   a=(72^2)/(0.445)=11649.44m/s^2

Centripetal acceleration = 11649.44m/s²

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answers

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=(1)/(2) kx^(2) +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

(1)/(2) kx^(2) +PE1=PE2

PE2-PE1=(1)/(2) kx^(2)

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

mgh=(1)/(2) kx^(2)

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

2100*9.8*h=(1)/(2)*2200*0.11^(2)

20580*h=13.31

h=(13.31)/(20580)

⇒h=0.0006467m=6.5e-4

Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off

Answers

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

\to (\$60)/((30 \ days)/(24\ hours)) = \$0.08 / kwh.

Thus, (\$0.08)/(\$0.12) = 0.694 \ kW *  0.694 \ kW  * 1000 = 694 \ W.

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

\to I = (P)/(V)= (694\ W)/(120\ V) = 5.78\ A

Energy usage reduction percentage = ((60\ W)/(694\ W) * 100\%)

This bulb accounts for 8.64\% of the energy used, hence it saves when you switch it off.

In a solution such as salt water, the component that does the dissolving is called the?

Answers

Answer:

Solvent

Explanation:

  • A solution is a substance that is made by dissolving one component in another.
  • It is made up of a solvent and a solute.
  • The component that dissolves the other is known as solvent. Examples of solvents include water, ethanol, milk, chloroform, etc.
  • The component that dissolves in the other is known as the solute. Examples of solute include salts, sugar, etc.
  • For example in an aqueous solution of sodium chloride(NaCl); sodium chloride salt is the solute while water is the solvent.

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?

Answers

The linear speed of the ball for the circular motion is determined as 12 m/s.

The given parameters;

  • mass of each ball, m = 450 g = 0.45 kg
  • length of the rod, L = 0.2 m
  • radius of the rod, r = 0.1 m
  • angular speed of the ball, ω = 120 rad/s

The linear speed of the ball is calculated as follows;

v = ωr

where;

  • ω is the angular speed of the ball
  • r is the radius of circular motion of the ball

The linear speed of the ball is calculated as follows;

v = ωr

v = 120 x 0.1

v = 12 m/s

Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.

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Answer:

The speed of ball is 12 (m)/(s)

Explanation:

Given:

Mass of ball m = 0.45 kg

Radius of rotation r = 0.1 m

Angular speed \omega = 120 (rad)/(s)

Here barbell spins around a pivot at its center and barbell consists of two small balls,

From the formula of speed in terms of angular speed,

  v  = r \omega

Where v = speed of ball

  v = 120 * 0.1

  v = 12 (m)/(s)

Therefore, the speed of ball is 12 (m)/(s)