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miss Akunina [59]
3 years ago
13

Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50.0 m and moves with a constant

speed of 6.5 m/s; motorcycle 2 starts from rest at x = 0 and moves with constant acceleration. Motorcycle 2 passes motor cycle 1 at the time t = 10.0s. What is the speed of motorcycle 2 when it passes motorcycle 1?
Physics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

Vf = 23 m/s

Explanation:

First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:

s₁ = v₁t₁

where,

s₁ = distance covered by motorcycle 1 = ?

v₁ = speed of motorcycle 1 = 6.5 m/s

t₁ = time = 10 s

Therefore,

s₁ = (6.5 m/s)(10 s)

s₁ = 65 m

Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,

s₂ = s₁ + 50 m

s₂ = 65 m + 50 m

s₂ = 115 m

Now, using second equation of motion for motorcycle 2:

s₂ = Vi t + (1/2)at²

where,

Vi = initial velocity of motorcycle 2 = 0 m/s

Therefore,

115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²

a = 230 m/100 s²

a = 2.3 m/s²

Now, using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (2.3 m/s²)(10 s)

<u>Vf = 23 m/s</u>

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Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
Match each property with the benefit it provides to people on Earth. A. magnetic field filters cancer-causing rays from the sun
vodomira [7]

The correct matching are the following:

  • A. magnetic field: prevents charged particles from the sun from reaching the surface
  • B. ozone: filters cancer-causing rays from the sun
  • C. carbon dioxide: retains heat energy from the sun in the atmosphere
  • D. water in oceans, lakes, rivers and streams: moderates changes in temperature on the surface

The electromagnetic field of the Earth protects us from solar wind by deflecting it. Without it, the charged particles would strip away our ozone, which will lead to the end of life on Earth.

The ozone is a layer in the stratosphere that filters the suns rays. It filters the UV rays by absorbing it. This prevents the UV rays from damaging the surface of the Earth.

Carbon dioxide retains heat in the atmosphere, which creates a greenhouse effect. It is beneficial for us, but because human activity like industrial activities emit so much carbon in the atmosphere that it ends up being harmful.

Bodies of water found on Earth regulate the temperature of the Earth through ocean currents. It brings in both cold and warm air to land which also affect precipitation.

Long explanation, I know, but maybe this will help you out in the long run. Good Luck!

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3 years ago
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A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of
Sholpan [36]
F=ma
where:
F - force
m - mass
a - acceleration 

We transform this formula to get a:
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a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
4 0
3 years ago
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The world's fastest production sportscar has a top speed of 415 kmh-1(a)Convert this speed to ms-1.[ 1](b)The distance from Lond
Alex73 [517]

Answer:

(a). The speed is 115.28 m/s.

(b). The time is 94 min.

(c). The International Space Station travels 43202.400 km.

(d). Speed is scalar quantity.

Velocity is vector quantity.

Explanation:

Given that,

Top speed = 415 km/h

1 miles = 1609 m

(a). We need to calculate the speed in m/s

Using conversion of km/h to m/s

v=415\ km/h

v=415\times\dfrac{5}{18}

v=115.28\ m/s

(b). We need to calculate the distance from London to Edinburgh in km

Using conversion for distance

d=403\ miles

Distance in meter,

d=403\times1609

d=648427\ m

d=648.427\ km

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{648.427}{415}

t=1.56\ h

t=1\ hours 34\ min

t= 94\ min

(c). Speed v' =7.66 km/s

v'=7660 m/s

In the time it takes the car to travel from London to Edinburgh,

We need to calculate the distance

Using formula of distance

d'=v'\times t

Put the value into the formula

d'=7660\times94\times60

d'=43202400\ m

d'=43202.400\ km

(d). Speed :

Speed is equal to the distance divided by time.

It is scalar quantity.

Velocity :

Velocity is equal to the displacement divided by time.

It is vector quantity.

Hence, (a). The speed is 115.28 m/s.

(b). The time is 94 min.

(c). The International Space Station travels 43202.400 km.

(d). Speed is scalar quantity.

Velocity is vector quantity.

3 0
2 years ago
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elixir [45]

Answer:

The horizontal displacement of the arrow is not larger than the banana split.

Explanation:

Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.

So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.

y - y₀ = ut - 1/2gt²

0 - y₀ = 0 × t - 1/2gt²

-y₀ = -1/2gt²

t² = 2y₀/g

t = √(2y₀/g)

Substituting the values of the variables, we have

t = √(2y₀/g)

= √(2 × 8848.04 m/9.8 m/s²)

= √(17696.08 m/9.8 m/s²)

= √(1805.72 s²)

= 42.5 s

The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s

So, d = vt

= 100 m/s × 42.5 s

= 4250 m

= 4.25 km

Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.

8 0
2 years ago
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