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luda_lava [24]
3 years ago
15

The hypotenuse of an isosceles right triangle is 36 units. to the nearest tenth, what is the length of one of the legs

Physics
1 answer:
GarryVolchara [31]3 years ago
3 0
The length of one of them would be 6
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Which of the following statements accurately describes the properties of gases?
Ray Of Light [21]
<span>A change in the pressure of a gas results in a more significant change in volume than it would in a liquid.  is the statement that accurately describes the property of gas. Gas only depends on how you store it. the bigger the space the wider gas can expand, the smaller the space, the more compress the gas can become.</span>
4 0
3 years ago
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What is the velocity of a car that traveled a total of 100 kilometers south in 2.5 hours?
lesya [120]
Using the equation v(average)=x traveled/time
v = 100/2.5
You get 40 kilometers per hour

Hope this helped!
7 0
3 years ago
The displacement vector from your house to the library is 520 m long, pointing 40 ∘ north of east.What are the x-component (x-ax
Alexeev081 [22]

Answer:

398.3 m, 334.2 m

Explanation:

The magnitude of the displacement vector is

v = 520 m

And its direction is

\theta=40^{\circ}

measured as north of east.

The x-component of this vector is given by:

v_x = v_0 cos \theta = (520)cos 40^{\circ}=398.3 m

While the y-component is given by

v_y = v_0 sin \theta =(520)sin 40^{\circ}=334.2 m

7 0
3 years ago
Which scientist is often called the “father of the atomic bomb” because of his work as the head of the manhattan project?
kirza4 [7]

Answer:

J. Robert Oppenheimer

Explanation:

He led the Manhattan project and created the first nuclear bomb in WWII

8 0
3 years ago
An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a In turn it forms an α-angle α
xxMikexx [17]
Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta -  \alpha ) \frac{d}{u} - h = 0
4.9[ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha }  ]^{2} - v_{0} sin(\theta -  \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha } ] - X sin \alpha  = 0
Hence obtain
aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta -  \alpha )}]^{2} \\  b = cos \alpha \,  tan(\theta -  \alpha ) + sin \alpha
The non-triial solution for X is
X= \frac{b}{a}

Answer:
X= \frac{sin \alpha  + cos \alpha  \, tan(\theta -  \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta -  \alpha )}  ]^{2}}

Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

Answer:  X = 101.5 m

7 0
3 years ago
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