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2 years ago

14

Which dog has the most kinetic energy? A. A dog of mass 12 kg running with speed 6 m/s B. A dog of mass 10 kg running with speed

3 m/s C. A dog of mass 16 kg running with speed 5 m/s D. A dog of mass 14 kg running with speed 4 m/s

1 answer:

Nonamiya [84]2 years ago

5 0

Answer:

A

Explanation:

KE = 1/2 m v^2

the greatest is A

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In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward

11111nata11111 [884]

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

$H=\dfrac{u^2\ sin^2\theta}{g}$

$10.9=\dfrac{u^2\ sin^2(63)}{9.8}$

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

4 0

3 years ago

13 points and brainlyest if possible. Thanks.

nikdorinn [45]

Most likely it would be C not completely sure

3 0

3 years ago

Read 2 more answers

A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass of 1.56 kg. Calculate it

Kazeer [188]

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = $\frac{m}{V}$

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = $\frac{1,56Kg}{1,767145L}$ = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) / φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

3 0

3 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

3 years ago

A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co

lidiya [134]

The position vector can be transcribed as:

A<span> = 6 i + y j </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102 = (6 i + y j)•(6 i+ y j) Note that i•j = 0, and i•i = j•j = 1 </span>

<span>100 = 36 + y2 </span>

<span>64 = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

3 0

3 years ago