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3 years ago

If the car went 30 km west in 25 min. and then 40 km south in 35 min., what would be its average speed?

Physics

2 answers:

Alika [10]3 years ago

8 0

The total distance is 70km.
The total time is 60 minutes or 1 hour.
Speed=Distance÷Time
=70÷1
=70km/h

gavmur [86]3 years ago

8 0

Average speed = (total-distance)/(total-time). That's 70 km/hr.

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Carbon has six protons. Which model shows a neutral atom of carbon?

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Answer:

Explanation:

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2 years ago

Problem 3) Bob stands at the edge of the swimming pool holding a laser 1.5m above the ground. He shines the red laser beam onto

Vadim26 [7]

Answer:

d = 5.75m

Explanation:

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

Here,

i = 90 - θ

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$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

$r = \sin^{-1}\frac{(1)\sin (90-26.56)}{1.33}\\\\r = 42.26m$

$\tan r = \frac{2.5}{d_1}$

$d_1 = \frac{2.5}{\tan (42.26)} \\\\d_1 = 2.75m$

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d = 3 + d₁

d = 3 + 2.75

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3 years ago

What is the maximum m2 value that the system can be stationary at?

Ne4ueva [31]

Answer: 37.5 kg in 3 s.f.

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3 years ago

F all of the energy in a falling object's gravitational potential energy store is transferred to its kinetic energy store by the

stepladder [879]

Answer:

The options are not shown, so let's derive the relationship.

For an object that is at a height H above the ground, and is not moving, the potential energy will be:

U = m*g*H

where m is the mass of the object, and g is the gravitational acceleration.

Now, the kinetic energy of an object can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.

Uinitial = Kfinal.

m*g*H = (1/2)*m*v^2

v^2 = 2*g*H

v = √(2*g*H)

So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.

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2 years ago

A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r

sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana $m_{b}$ = 8.97 kg

at distance r = 4/5 { radius of platform }

mass of monkey $m_{m}$ = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + $m_{b}$ ( $\frac{4}{5}$ R)² + $m_{m}$ R² ]w

m/2 × ω₀ = [ m/2 + $m_{b}$ ( $\frac{4}{5}$ )² + $m_{m}$ ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

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3 years ago