A 10-cm-long archerfish begins such a leap with its body vertical in the water, using its fins and undulating its body to propel itself sharply upward. Its acceleration steadily decreases as it leaves the water because less of its body and fewer of its fins are in contact with the water.

With what speed does the fish leave the water?
What height above the water does it reach?

Answer:

(a) distance =27m

(b)acceleration =14/3ms-2

Explanation:

s=(u+v)t/2

= (2+16)3/2

=27m

v=u+at

16=2+a*3

a= 14/3ms-2

Answer:

Explanation:

ASSUMING the belt is horizontal

kinetic friction force is μmg = 0.6(8)(9.8) = 47.04 N

Horizontal acceleration is

a = F/m = 47.04 / 8 = 5.88 m/s²

t = v/a = 5.0 / 5.88 = 0.85034...

t = 0.85 s

Answer:

B. When an energy transformation happens, no energy is destroyed

or created.

Explanation:

Answer:

it will be curved as in deceleration

Explanation:

Answer:

Explanation:

Use vector addition

v = √(30² + 40²) = 50 m/s as viewed from the shore

Trust me when I say I would never venture onto a river moving at 30 m/s no matter how powerful the boat was.

30 m/s(3600 s/hr / 1000 m/km) = 108 km/hr

The turbulence in water flowing over ground at 108 kilometers per hour would capsize anything ever built...including aircraft carriers.

These numbers are more suitable for a plane flying in air. Even that could be a very rough ride.

Answer:

Answer:

b) the height the ball bounces

Explanation:

the control variable is the variable that you change yourself. since you change the height that the ball bounces from we know this is the answer

Answer:

Yeah it's the new trend lol it's kinda weird

Answer:

it is marked by mystery and ambiguity. If to this fascination for creating tension is added a feeling of fear

Explanation:

do I help you

Answer:

388.56km[E13degS]

Explanation:

Δdx = Δd1x + Δd2x

Δdx = 250km [E]+ 180sin(60) [E]

Δdx = 405.88 km [E]

Δdy = Δd1y + Δd2y

Δdy = 0 + 160cos(60) [S]

Δdy = 90km [S]

Δdt = √(Δdx)^2+(Δdy)^2

Δdt = √164738.57+8100

Δdt = 398.85km

tanθ = Δdy/Δdx

tanθ = 90/388.56

θ = tan-1(90/388.56)

θ = 13.04 deg

A large system is one in which one subsystem provides services to another subsystem.

This is a type of system that contains many subsystems. The subsystem provides independent services to the larger by communicating, or interacting with each other.

A subsystem is smaller units of a large system, designed to perform specific task in the large system.

Thus, a large system is one in which one subsystem provides services to another subsystem.

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im not sure if im right or not but maybe its "D scientific investigation" but ldk so

Answer:

2N/cm²

Explanation:

1m=100cm so, 1m²= 100×100 cm² = 10000cm².

therefore , 20000N/m²= 20000/10000 N/cm².

= 2N/cm².

hope this helps you.

Answer:

The mass of the gold bar is 1,544 g

Explanation:

Answer:

water and nitrogen can be place at B

The B loop is the element that can be liquid, cas or solid. Water can have three states : solid (ice), liquid, or gas (as a vapor). The same can be said with nitrogen, it's a gas at room temperature, and it also can be a liquid or solid.

Answer:

40 m/s

Explanation:

If you multiply the acceleration with time you get the average speed.

Hi there!

We know that:

I = Δp = mΔv = m(vf - vi)

Let the positive direction be towards the wall. Plug in the given values:

Δp = 0.4(-25 - 12) = -14.8 Ns

Δp = Ft, so:

Δp/t = F

-14.8/0.05 = -296 N (depending on which direction you assign positive/negative, the answer would be positive/negative)

Answer:

[tex]\boxed {\boxed {\sf 2.2 \ m/s^2}}[/tex]

Explanation:

We are asked to solve for the magnitude of the car's acceleration.

We are given the initial speed, final speed, and distance, so we will use the following kinematic equation.

[tex]{v_f}^2={v_i}^2+2ad[/tex]

The car is initially traveling at 15.0 meters per second and accelerates to 20.0 meters per second over a distance of 40.0 meters. Therefore,

Substitute the values into the formula.

[tex](20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)[/tex]

Solve the exponents.

[tex]400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)[/tex]

Subtract 225.0 m²/s² from both sides of the equation.

[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 225.0 \ m^2/s^2 -225 \ m^2/s^2 +2a(40.0 \ m)[/tex]

[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)[/tex]

[tex]175 \ m^2/s^2 = 2a(40.0 \ m)[/tex]

Multiply on the right side of the equation.

[tex]175 \ m^2/s^2 =80.0 \ m *a[/tex]

Divide both sides by 80.0 meters to isolate the variable a.

[tex]\frac {175 \ m^2/s^2}{80.0 \ m}= \frac{80.0 \ m *a}{80.0 \ m}[/tex]

[tex]\frac {175 \ m^2/s^2}{80.0 \ m}=a[/tex]

[tex]2.1875 \ m/s^2 =a[/tex]

Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.

[tex]2.2 \ m/s^2=a[/tex]

The magnitude of the car's acceleration is 2.2 meters per second squared.

(1) The speed with which the star's light will pass you is 3 x 10⁸ m/s.

(2) The speed of the other spaceship is -2.943 x 10⁸ m/s.

The given parameters;

(1)

Since you are travelling away from the star (same direction), relative to a third observer, the speed at which the star's light will pass you is calculated using Lorentz velocity addition.

[tex]u = \frac{u'+ v}{1 + \frac{v}{c^2} u'} \\\\u = \frac{c+(-0.8c)}{1 + \frac{-0.8c}{c^2} (c)}\\\\u = \frac{c-0.8c}{1- \frac{0.8c}{c} } \\\\u = \frac{0.2c}{0.2} = c = 3\times 10^{8} \ m/s[/tex]

Thus, the speed with which the star's light will pass you is 3 x 10⁸ m/s.

(2) The speed of the spaceship approaching = 0.84 c

Since you are travelling in opposite direction to second spaceship, relative to a third observer (frame of reference), the speed of the other spaceship is calculated using Lorentz velocity addition.

[tex]u = \frac{u' + v}{1 + \frac{v}{c^2}u' } \\\\u = \frac{-0.84c - 0.8c}{1 + \frac{(-0.8c)}{c^2}(-0.84c) } \\\\u = \frac{-1.64c}{1.672} = -0.981c \\\\u = -0.981 \times 3\times 10^{8} = -2.943 \times 10^{8} \ m/s[/tex]

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Answer:

m = 4 At Mass units = 4 * 1.66E-27 kg = 6.64E-27 kg  mass of particle

q = 2 * 1.60E-19 C = 3.2E-19 C     total charge of alpha particle

T = 8.8E-8 s     period of 1 revolution

F = q V B      magnetic force

F = m V^2/ R     centripetal acceleration

q V B = m V^2 / R

B = m V / (q * R)

T = 2 pi R / V     time for 1 revolution

R = V T / (2 pi)

B = (m V / q) * 2 pi / (V T) = 2 pi m / (q T)

B = 2 pi * 6.64E-27 / (3.2E-19 * 8.8E-8) = 1.48

B = 1.48 W/m^2

Answer:

π/21600 rad /s.

Answer:

c 13

Explanation:

I took test

Group 18 of the periodic table, also known as the noble gases or Group 18 elements, consists of elements that share similar properties and have a single electron in their outer shells. The correct option is option (A).

These elements include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). They are characterized by having full outer electron shells, making them stable and unreactive under normal conditions.

This stability is due to the fact that they have achieved a full complement of electrons in their outermost energy level, except for helium, which has only two electrons in total.

Group 18 of the periodic table, also known as the noble gases or Group 18 elements, The correct option is option (A).

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Answer:

60

Explanation:

48-(-12)=60

the -(- is like a big plus sign really if you see that then just add the numbers together so do 48 plus 12 it equals 60.

Answer:

f

Explanation:

The speed of the two-car wreck right after the crash is 13.9 m/s, and the direction is 58.4° east of north.

Momentum conservation can address this problem. Before and after the collision, momentum is equal. Calculate the two-car wreck's speed and direction after the collision:

Step 1: Calculate automobile beginning momentum.

Mass (m) times velocity (v) equals momentum (p).

Northbound 1200-kg car:

Initial momentum of the 1200-kg automobile (p₁) = mass (m₁) × velocity (v₁).

p₁ = 1200 kg × 17.0 m/s = 20,400 kg/s (North direction).

1600-kg eastbound car:

The 1600-kg car's initial momentum (p₂) = m₂ × v₂.

p₂ = 1600 kg × 21.0 m/s = 33,600 kg/s (East direction).

Step 2: Calculate the pre-collision momentum.

The vector sum of the two automobiles' momentum before the accident is p[tex]_total[/tex].

p[tex]_total[/tex] = p₁+p₂.

p[tex]_total[/tex] = 20,400 kg m/s (North) + 33,600 (East)

Step 3: Calculate the two cars' combined mass after the collision.

Combined mass (m[tex]_combined[/tex]) = 1200-kg + 1600-kg cars.

m[tex]_combined[/tex] = 1200 + 1600 = 2800 kg

Step 4: Determine the two-car wreck's velocity after the accident.

Total momentum after the collision (p[tex]_combined[/tex]) is the same as before, but it belongs to the combined mass (m[tex]_combined[/tex]) travelling at the velocity (v[tex]_combined[/tex]) in a given direction:

p[tex]_combined[/tex] = m[tex]_combined[/tex] × v[tex]_combined[/tex]

Since the collision is inelastic (cars stick together), momentum remains the same.

= p[tex]_total[/tex]

m[tex]_combined[/tex] × v[tex]_combined[/tex] = 20,400 kg/s (North) + 33,600 kg/s (East).

Step 5: Determine the two-car wreck's speed and direction.

Calculating the resultant vector of the North and East momentum components gives the combined two-car wreck's velocity (speed) (v[tex]_combined[/tex]).

(v[tex]_combined[/tex][tex]_north²[/tex] + v[tex]_combined[/tex][tex]_east²[/tex])

Where:

North velocity is v[tex]_combined[/tex][tex]_north[/tex].

East velocity is v[tex]_combined[/tex][tex]_east[/tex].

v[tex]_combined[/tex][tex]_north[/tex] = 20,400 kg m/s, 2800 kg, 7.29 m/s (North).

v[tex]_combined[/tex][tex]_east[/tex] = 33,600 kg m/s < 2800 kg × 12.0 m/s (East).

13.9 m/s = (7.29² + 12.0²)

Step 6: Determine the two-car wreck's direction.

Trigonometry can determine direction:

tan() = v[tex]_combined[/tex][tex]_east/north[/tex]

arctan(v[tex]_combined[/tex][tex]_east/north[/tex]) =

arctan(12.0 / 7.29 m/s) arctan(1.645) 58.4°

The two-car crash faces east at 58.4°.

Results summary:

The two-car disaster shortly after the accident is 13.9 m/s.

The two-car crash lies 58.4° east of north following the collision.

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Answer: (a) Write position-time equations for both cars. Let east be the positive direction.

Explanation:

Hope this helps!

Answer:

the bus travles 65x7=455 km

Explanation:

hope this helps you

Answer:

455 km/h

Explanation:

65km/h x 7 hours = 455 km/h

Answer:

is it a poem or recipe :(

don't know

Answer:

Explanation:

Conservation of momentum is used to solve

Unfortunately we have a missing piece of information such as the initial velocity of the unknown mass train.

If we ASSUME that the second train is at rest

5000(100) + m(0) = 5000(50) + m(50)

which means m = 5000 kg

However, I'll show you the importance of knowing that initial velocity by finding it assuming the other answers are valid

if m = 15000 kg

5000(100) + 15000(v₀) = (5000 + 15000)(50)

v₀ = 33 ⅓ m/s

if m = 10000 kg

5000(100) + 10000(v₀) = (5000 + 10000)(50)

v₀ = 25 m/s

if m = 8000 kg

5000(100) + 8000(v₀) = (5000 + 8000)(50)

v₀ = 18.75 m/s

So you can see why I had to assume an initial velocity. Any of the masses could work if the initial velocity is chosen correctly.

If you look at the sketch I drew for the earlier part of this question, you'll see that, with respect to the positive horizontal (i.e. directly to the right), T₁ makes an angle of 180° - 54.9205° ≈ 125°, while T₂ makes an angle of 54.9205° ≈ 55°.

Split up the force acting on the block into vertical and horizontal components. We have

• net vertical force

∑ F = T₁ sin(125°) + T₂ sin(55°) - mg = 0

where m = 0.56 kg, and

• net horizontal force

∑ F = T₁ cos(125°) + T₂ cos(55°) = 0

Both net forces are 0 because the block is suspended in equilibrium.

Notice that cos(125°) = -cos(55°), so the second equation tells you that T₁ = T₂ and that the tensions in either string are the same. Also, sin(125°) = sin(55°).

Then in the first equation, we have

T₁ sin(125°) + T₂ sin(55°) - mg = 0

2 T₁ sin(55°) = mg

T₁ = mg/(2 sin(55°))

T₁ = (0.56 kg) (9.8 m/s²)/(2 sin(55°))

T₁ ≈ 3.35 N