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Vlad1618 [11]
3 years ago
15

1.2miles=__________km

Physics
2 answers:
ioda3 years ago
7 0

Answer:

1.931 kilometres is the answer of 1.2 miles

larisa [96]3 years ago
7 0

Answer and Explanation:

1 mile = 1.609 km

Set up a fraction to cancel the miles to get the kilometers.

\frac{1.2mi}{?km} *\frac{1.609}{1mi} = 1.9308km<u> <- This is the answer.</u>

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
Verizon [17]

Answer:29.41 m/s

Explanation:

Given

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Height of river bank=2.4 m

width of river=60.0 m

Car can considered as projectile

Time travel to cover vertical distance=22.8-2.4=20.4 m

h=ut+\frac{at^2}{2}

20.4=0+\frac{gt^2}{2}

t^2=4.163

t=2.04 s

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Let u be the speed of car

ut=60

u=\frac{60}{2.04}=29.41 m/s

4 0
3 years ago
How much KCI can be dissolved in 100 g of water at 30°C?​
asambeis [7]

Answer:

34

Explanation:

The problem provides you with thesolubility of potassium chloride, KCl , in water at 20∘C , which is said to be equal to 34 g / 100 g H2O . This means that at 20∘C , a saturated solution of potassium chloride will contain 34 g of dissolved salt for every100 g of water.

6 0
3 years ago
Read 2 more answers
A potter's wheel has the shape of a solid uniform disk of mass 13.0 kg and radius 1.25 m. It spins about an axis perpendicular t
Fittoniya [83]

Answer:10.82 kg-m^2

Explanation:

Given

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mass of lump m=1.7 kg

distance of lump from axis r_0=0.63

Moment of inertia is the distribution of mass from the axis of rotation

Initial moment of inertia of disk I_1=\frac{Mr^2}{2}

I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2

Final moment of inertia I_f=Moment of inertia of disk+moment of inertia of lump about axis

I_f=\frac{Mr^2}{2}+mr_0^2

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3 0
4 years ago
43. A rocket sled accelerates at a rate of 49.0m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component
lord [1]

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

F_x = m a_x

where

F_x is the horizontal component of the force

m is the mass of the passenger

a_x is the horizontal component of the acceleration

Here we have

m = 75.0 kg

a_x = 49.0 m/s^2

Substituting,

F_x=(75.0)(49.0)=3675 N

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

R=mg=(75.0)(9.8)=735 N

where we used

g=9.8 m/s^2 as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

F_y = 735 N

and therefore the magnitude of the net force is

F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N

And the direction is given by

\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}

4 0
3 years ago
Solve for the final velocity using the picture below what type of collision is this an example of
mariarad [96]

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