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Veseljchak [2.6K]
3 years ago
12

A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.

1 s interval immediately after braking begins, the speed decreases to 15 m/s. What distance does the motorcycle travel from the instant braking begins until the motorcycle stops
Physics
1 answer:
shutvik [7]3 years ago
5 0

Answer:

The motorcycle travelled 69.73 m during these 3.1 s.

Explanation:

In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:

a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2

The distance is given by the following formula:

S = (V_final^2 - V_initial^2)/(2*a)

S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m

The motorcycle travelled 69.73 m during these 3.1 s.

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How can you justify that force is transferred to lift and throw soil using shovel?<br>​
Natali [406]

Forces are needed to lift, turn, move, open, close, push, pull, and so on. When you throw a ball, you are using force to make the ball move through the air. More than one force can act on an object at the same time.

4 0
2 years ago
Solution A has a specific heat of 2.0 J/g◦C. Solution B has a specific heat of 3.8 J/g◦C. If equal masses of both solutions start
fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

  • \Delta T= temperature difference
  • Q= heat energy
  • m= mass of the body
  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
  • equal heat is supplied to both the solutions, i.e. Q_A=Q_B
  • specific heat of solution A, c_{A}=2.0 J.g^{-1} .\degree C^{-1}
  • specific heat of solution B, c_{B}=3.8 J.g^{-1} .\degree C^{-1}
  • \Delta T_A & \Delta T_B are the change in temperatures of the respective solutions.

Now, putting the above values

Q_A=Q_B

m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1

Which proves that solution A attains a higher temperature than solution B.

7 0
3 years ago
The speed of light (c) is:
Radda [10]

Answer:

A

Explanation:

Speed of light is 299 792 458 m / s. So option A is answer.

7 0
2 years ago
A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find
____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s)  u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

 t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t  = time

           θ = 1 rev.

Since 1 rev = 2π (rad)

           t = 1.37 s

 Average angular velocity = 2π/t

π = 3.143

 Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

   Average angular velocity ≈ 4.59 rad/s

8 0
3 years ago
Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
3 years ago
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