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Maurinko [17]
3 years ago
8

A car travels 35 km west and 90 km north in two hours what is its average velocity?

Physics
1 answer:
blagie [28]3 years ago
8 0
Average velocity =  Total Displacement / Total time taken

Total displacement is the distance from starting to end point.

35km west then 90km north. When you will draw this and connect starting to end point. You will get a right angled triangle. 

So displacement = \sqrt{35^2+90^2}

= 96.56 km

So, Total displacement = 96.56 km

So, Average Velocity = 96.56 / 2  = 48.28 km/hr
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Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner t=\frac{295.6}{4.65}=63.56 s

time required by first runner to reach finish line=\frac{250}{3.1}=80.64 s

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0
4 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
4 years ago
What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest w
devlian [24]

Answer:

448 m/s is the correct answer.

Explanation:

7 0
4 years ago
Read 2 more answers
What is the resistance at 20°C of a 2.0-meter length of tungsten wire with a cross-sectional area of 7.9 10^-7
Bad White [126]

Answer:

1.4 * 10 ^-1 Ω

Explanation:

Hi,

For this question, we gotta use the formula

R = pL/A

p = The resistivity of your material at 20°C

L = length of the wire

A = cross-sectional area

The resistivity of tungsten is 5.60 * 10^-8 at 20°C

By plugging the values, we get:

R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω

8 0
3 years ago
Carol drops a stone in a mine shaft 122.5 metres deep.How
Verizon [17]

Answer:

 T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken  = ?

time taken by the stone to reach at the bottom

using equation of motion

s = u t + \dfrac{1}{2}gt^2

initial speed , u = 0 m/s

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 122.5}{9.8}}

       t = 5 s

time taken by the sound to travel

    d =v x t

 t = \dfrac{d}{v}

 t = \dfrac{122.5}{340}

    t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

 T = 5.36 s

7 0
3 years ago
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