All categories

2 years ago

Annie is considering two different coach journeys from Bristol to Birmingham.

Physics

1 answer:

Alexeev081 [22]2 years ago

7 0

The average speed in the first case is 19.86 m/s and the average speed in the second case is 13.67 m/s.

The formula of average speed is

average speed = total distance / total time

In first case

total distance is 143 km which is also 143000 m

and total time is 2 hours which is also 7200 seconds

so average speed = ( 143000 / 7200 ) m/s

average speed = 19.86 m/s

In second case

total distance is 160 km which is also 160000 m

and total time is 3 hours and 15 minutes which is also 11700 seconds

so average speed = ( 160000 / 11700 ) m/s

average speed = 13.67 m/s

So to conclude with we have find the the average speed in both cases by dividing total distance by total time and in first case we have got our answer as 19.86 m/s and in second case we have got 13.67 m/s.

Learn more about speed here:

brainly.com/question/4931057

#SPJ10

You might be interested in

40 POINTS <br> Can someone please answer me questions 16 and 17

Tatiana [17]

16. Solids cannot transport heat through convection. Because solids are stuck in place so they can't flow and since they can't flow, there is no convection.
17. It is because the metal conducts heat faster that it feels colder than the wood, which conducted heat slower. In other words metals are good conductors of heat. It takes away your body heat as you touch it and which makes your body temperature to drop, therefore you feel the coldness when you touch a piece of metal.

6 0

3 years ago

A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through

aniked [119]

Answer:

$a_2=24.5\ \text{m/s}^2$ towards right

$a_1=2.1875\ \text{m/s}^2$ towards left

Explanation:

$m_1$ = Mass of skater = 56 kg

$m_2$ = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2$

So, the ball will move towards right with a magnitude of acceleration $a_2=24.5\ \text{m/s}^2$ .

The force on the ball will be

$F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}$

The force on the ball is $122.5\ \text{N}$

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

$-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2$

So, the skater will move towards left with a magnitude of acceleration $2.1875\ \text{m/s}^2$

4 0

3 years ago

One ball of mass 0.600kg travelling 9.00m/s to the right collides head on elastically with a second ball of mass 0.300kg travell

Alina [70]

Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

where v₁' and v₂' are the balls' respective velocities after the collision.

Kinetic energy is also conserved, so

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²

m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²

From the momentum equation, we have

(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'

which simplifies to

10.0 m/s = 2 v₁' + v₂'

so that

v₂' = 10.0 m/s - 2 v₁'

From the energy equation, we have

(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

which simplifies to

67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

226 m²/s² = 2 (v₁')² + (v₂')²

Substituting v₂' yields

226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²

which simplifies to

3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0

Solving for v₁' using the quadratic formula gives two solutions,

v₁' ≈ -2.33 m/s or v₁' = 9.00 m/s

but the second solution corresponds to the initial conditions, so we omit that one.

Then the second ball has velocity

v₂' = 10.0 m/s - 2 (-2.33 m/s)

v₂' ≈ 14.7 m/s

6 0

2 years ago

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

Sedaia [141]

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0

3 years ago

India used 600 joules of work on a steel block and moved it 4.8 meters. How much forced was used ?

dangina [55]

The formula for finding force is F=W/d so then you plug in the numbers:
F=600/4.8
F=125N

8 0

3 years ago