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2 years ago

In which graph is acceleration the slope?

Physics

2 answers:

shtirl [24]2 years ago

7 0

Answer:

it may help u I hope leave comment plz

olga2289 [7]2 years ago

4 0

Answer:

Velocity -time graph 

Explanation:

hope it helps ✌️✌️

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If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:

$\theta = 1.22\frac{\lambda}{d}$

Where,

$\lambda$ = Wavelength

d = Width of the slit

$\theta$ = Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

$r = l\theta$

Relacing $\theta$

$r = l(\frac{1.22\lambda}{d})$

$r = 1.22\frac{l\lambda}{d}$

Replacing with our values we have that,

$r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})$

$r = 3680.91m$

Therefore the diameter of the beam on the moon is

$d = 2r$

$d = 2 * (3690.91)$

$d = 7361.8285m$

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0

3 years ago

Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), th

Zinaida [17]

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

5 0

3 years ago

A regulation basketball has a 32 cm diameter

Rudik [331]

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

3 0

3 years ago

A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?

iris [78.8K]

Answer:

$\Delta x=22.67786838m$

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

$y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2$ (1)

Where:

$y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$v_o=initial\hspace{3}velocity$

$t=travel\hspace{3}time$

$g=gravity\hspace{3}constant$

$\theta=Initial\hspace{3}launch\hspace{3}angle$

In this case:

$\theta=0$

Because the dog jumps horizontally

Let's asume the gravity constant as:

$g=9.8$

$y=0$

Because when the dog reach the base the height is 0

$y_o=70$

$v_o=6$

Now let's replace the data in (1)

$y_o-\frac{1}{2} *(9.8)*t^2+70$

Isolating t:

$t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473$

Finally let's find the horizontal displacement using this equation:

$\Delta x=v_o*cos(\theta)*t$

Replacing the data:

$\Delta x=6*1*3.77964473=22.67786838m$

8 0

3 years ago

A 1,600 kg car traveling north at 10.0 m/s crashes into a 1,400 kg car traveling east at 15 m/s at an unexpectedly icy intersect

tigry1 [53]

Here it is the use of vector and conservation of momentum !

so,
√(16^2+21^2) ×1000= 3000 v
v =8.8 m/s

so answer is B !

if you have any doubt, you can ask ! just comment !

7 0

3 years ago